Question:

Integrate the function x+3/x^2−2x−5

Answer:

1. Rewrite the function in terms of the derivative of x: x + 3/x^2 - 2x - 5 = (1/x^2 - 2)dx

2. Integrate both sides: ∫(1/x^2 - 2)dx = ∫(x + 3/x^2 - 2x - 5)dx

3. Solve the left side of the equation: -1/x + 2x = ∫(1/x^2 - 2)dx

4. Solve the right side of the equation: -1/x + 2x = ∫(x + 3/x^2 - 2x - 5)dx

5. Combine the two equations: ∫(1/x^2 - 2)dx = ∫(x + 3/x^2 - 2x - 5)dx = -1/x + 2x

Question:

∫xdx​/(x−1)(x−2) equals log∣​(x−1)^2​/x−2∣+C log∣(x−2)^2​/x−1∣​+C log∣​(x−1​/x−2)^2∣+C log∣(x−1)(x−2)∣+C

Answer:

Answer:

∫xdx/(x−1)(x−2) = ∫xdx/(x-1) - ∫xdx/(x-2)

= ∫xdx/(x-1) - ∫(x-1)dx/(x-2)

= log|x-1| - log|x-2| + C

= log|(x-1)/(x-2)| + C

= log|(x-1)(x-2)| + C

Question:

Integrate the rational function 2x/x^2+3x+2

Answer:

Step 1: Rewrite the rational function as (2x)/(x^2 + 3x + 2)

Step 2: Factor the denominator of the rational function as (x + 2)(x + 1)

Step 3: Rewrite the rational function as (2x)/((x + 2)(x + 1))

Step 4: Use the formula for integration of rational functions: ∫(2x)/((x + 2)(x + 1))dx = (1/2)ln|x + 2| - (1/2)ln|x + 1| + C

Question:

Integrate the function x+2/√4x−x^2

Answer:

Step 1: Rewrite the function in standard form by multiplying both numerator and denominator by the conjugate of the denominator.

x + 2/√4x - x^2 = (x + 2)(√4x + x^2)/(4x - x^4)

Step 2: Use the formula for integration of a rational function to solve the integral.

Integral (x + 2)(√4x + x^2)/(4x - x^4) dx = (1/2)∫(√4x + x^2)dx - (1/4)∫(x^2 + 2x)dx

Step 3: Use the power rule to solve the integrals.

(1/2)∫(√4x + x^2)dx = (1/2)((2/3)x^(3/2) + (1/3)x^(3/2)) + C

(1/4)∫(x^2 + 2x)dx = (1/4)(x^3 + x^2) + C

Step 4: Combine the two integrals.

Integral (x + 2)(√4x + x^2)/(4x - x^4) dx = (1/2)((2/3)x^(3/2) + (1/3)x^(3/2)) + (1/4)(x^3 + x^2) + C

Question:

Integrate the rational function 3x−1/(x−1)(x−2)(x−3)

Answer:

1. Multiply the numerator and denominator of the rational function by the conjugate of the denominator:

(3x−1)(x−2)(x+3)/(x−1)²(x−2)(x−3)

2. Expand the numerator and denominator:

3x³-7x²+13x-3/ (x²-4x+3)(x−1)(x−2)(x−3)

3. Split the fraction into partial fractions:

A/x-1 + B/x-2 + C/x-3 + (3x²-7x+13)/(x²-4x+3)

4. Find the values of A, B, and C by using the following equations:

A(x²-4x+3) + B(x-1)(x-3) + C(x-1)(x-2) = 3x²-7x+13

A(x²-4x+3) = 3x²-7x+13

A = 1

B(x-1)(x-3) + C(x-1)(x-2) = 0

B(x-1)(x-3) = 0

B = 0

C(x-1)(x-2) = 0

C = 0

5. Substitute the values of A, B, and C into the partial fractions:

A/x-1 + B/x-2 + C/x-3 + (3x²-7x+13)/(x²-4x+3)

1/x-1 + 0/x-2 + 0/x-3 + (3x²-7x+13)/(x²-4x+3)

6. Integrate each term:

∫1/x-1 dx = ln|x-1| + C

∫0/x-2 dx = 0 + C

∫0/x-3 dx = 0 + C

∫(3x²-7x+13)/(x²-4x+3) dx = (1/2)ln|x²-4x+3| + C

7. Combine the terms:

ln|x-1| + 0 + 0 + (1/2)ln|x²-4x+3| + C

Question:

Integrate the rational function x/(x^2+1)(x−1)

Answer:

1. Rewrite the equation as:

(x/(x^2+1))*(x-1)

2. Use the product rule to integrate:

(1/(x^2+1))(x-1) + (x/(x^2+1))(1)

3. Simplify:

(1/(x^2+1))*(x-1) + (x/(x^2+1))

4. Integrate:

(1/2)*ln(x^2+1) - (1/2)*ln(x-1)

Question:

Integrate the function sec^2x/tan^2x+4

Answer:

1. Let u = tanx
2. du = sec^2x dx
3. ∫sec^2x/tan^2x+4 dx = ∫ (1/u^2+4)du
4. ∫sec^2x/tan^2x+4 dx = 1/4 ∫ (4/u^2+1)du
5. ∫sec^2x/tan^2x+4 dx = 1/4 ln|u^2+4| + C
6. ∫sec^2x/tan^2x+4 dx = 1/4 ln|tan^2x + 4| + C

Question:

∫1+x2 dx is equal to A 2x​1+x2+21​log∣+1+x2∣+C B 2​/3(1+x^2)^2​/3+C C 2​/3x(1+x^2)^3​/2+C D x^2​/2√1+x^2+1​/2x^2log∣x+1+x^2​∣+C

Answer:

A. 2x​1+x2+21​log∣+1+x2∣+C

Question:

Integrate the function 1/√x^2+2x+2

Answer:

Answer: Step 1: Separate the function into two parts: 1/√x and x^2+2x+2 Step 2: Take the integral of the first part: 2*arctan(√x) Step 3: Take the integral of the second part: (1/3)x^3 + x^2 + 2x Step 4: Add the two integrals together: 2arctan(√x) + (1/3)*x^3 + x^2 + 2x

Question:

Integrate the function √x^2+4x+6

Answer:

Answer: Step 1: Rewrite the function into a polynomial form. √x^2+4x+6 = (x+3)^2

Step 2: Use the power rule to integrate the polynomial. Integral of (x+3)^2 dx = (x+3)^3/3 + C

Question:

Integrate the function 1/x−√x

Answer:

1. ∫1/x−√x dx
2. ∫1/x dx - ∫√x dx
3. ln|x| - 2/3x^(3/2) + C

Question:

∫10x^9+10^xloge​10dx​/x^10+10^x equals A 10^x−x^10+C B 10^x+x^10+C C (10^x−x^10)^−1+C D log(10^x+x^10)+C

Answer:

Answer: A 10^x−x^10+C

Question:

Integrate the function 1​/x(logx)^m,x>0,m=1

Answer:

Answer: Step 1: Use integration by parts. Let u = ln(x) and dv = 1/x dx

Step 2: Calculate du and v du = 1/x dx v = -x^-m

Step 3: Calculate the integral Integral of 1/x (ln(x))^m dx = -x^-m ln(x) + m ∫x^-m+1 dx

Step 4: Solve the integral Integral of x^-m+1 dx = x^-m+1/(-m - 1) + C

Question:

Integrate the function x^3sin(tan^−1x^4)/1+x^8

Answer:

1. First, use the chain rule to rewrite the function:

u = tan⁻¹(x⁴) du = (1/1+x⁴²) dx

x³sin(u) du/dx = x³sin(tan⁻¹(x⁴))(1/1+x⁴²)

2. Now, integrate the function:

∫x³sin(tan⁻¹(x⁴))(1/1+x⁴²) dx

= ∫x³sin(u)(1/1+x⁴²)du

= -x³cos(u)/1+x⁴² + ∫(3x²cos(u) - x⁴sin(u))/1+x⁴² dx

3. Finally, use integration by parts to solve the integral:

= -x³cos(u)/1+x⁴² + 3x²sin(u)/1+x⁴² - ∫x²(-sin(u) + x²cos(u))/1+x⁴² dx

= -x³cos(u)/1+x⁴² + 3x²sin(u)/1+x⁴² - x²sin(u)/1+x⁴² + ∫x⁴cos(u)/1+x⁴² dx

= -x³cos(u)/1+x⁴² + 3x²sin(u)/1+x⁴² - x²sin(u)/1+x⁴² + x⁴cos(u)/2 + C

= -x³cos(tan⁻¹(x⁴))/1+x⁴² + 3x²sin(tan⁻¹(x⁴))/1+x⁴² - x²sin(tan⁻¹(x⁴))/1+x⁴² + x⁴cos(tan⁻¹(x⁴))/2 + C

Question:

Integrate the function xsin3x

Answer:

1. Integrate the function xsin3x

2. ∫xsin3xdx

3. ∫xsinx3dx

4. ∫x(sinx)(3cosx)dx

5. ∫x(sinx)dx + ∫x(3cosx)dx

6. ∫x(sinx)dx + 3∫(cosx)dx

7. (1/2)x²sinx + 3sinx + C

Question:

Integrate the function (x^2+1)logx

Answer:

Answer: Step 1: Use the formula for integration by parts: ∫(u*dv)=uv-∫vdu

Step 2: Let u = x2 + 1 and dv = logx dx

Step 3: Calculate du and v: du = 2x dx and v = x logx - x

Step 4: Substitute the values in the formula: ∫(x2+1)logx dx = (x2+1)(x logx - x) - ∫(2x)(x logx - x)dx

Step 5: Integrate the second part of the equation: ∫(x2+1)logx dx = (x2+1)(x logx - x) - (2x2x logx - x2 - 2∫x logx dx)

Step 6: Integrate the third part of the equation: ∫(x2+1)logx dx = (x2+1)(x logx - x) - (2x2x logx - x2 - 2(x logx2 - 2logx + C))

Step 7: Simplify the equation: ∫(x2+1)logx dx = x2+1)(x logx - x) - 2x2x logx + 2x2 + 2x logx2 - 4logx + C

Question:

Integrate the function 5x+3/√x^2+4x+10

Answer:

1. Rewrite the fraction as a single fraction: (5x + 3)(√x^2 + 10)/(√x^2 + 4x + 10)

2. Multiply the numerator and denominator by the conjugate of the denominator: (5x + 3)(x + 5)/x^2 + 10

3. Rewrite the numerator as a single fraction: (5x^2 + 15x + 15)/x^2 + 10

4. Use the substitution u = x^2 + 10: (5u - 5x + 15)/u

5. Integrate the function: (5/2)u^2/2 - (5/2)xu + 15u + C

Question:

Integrate the function √x^2+4x+1

Answer:

1. Rewrite the function as (x+2)^2: (x+2)^2

2. Use the formula for the antiderivative of a perfect square: ∫(x+2)^2 dx = (1/3)(x+2)^3 + C

3. Simplify the answer: (1/3)(x+2)^3 + C

Question:

∫ dx​/x^2+2x+2 equals A xtan^−1(x+1)+C B tan^−1(x+1)+C C (x+1)tan^−1x+C D tan^−1x+C

Answer:

Answer: B tan^−1(x+1)+C

Question:

Integrate the function sec^2(7−4x)

Answer:

1. Integrate sec^2(7−4x)

2. Integrate sec^2(u) where u = 7−4x

3. Integrate 1/cos^2(u) du

4. Integrate 1/cos(u) du

5. Let v = cos(u)

6. Integrate 1/v dv

7. ∫ 1/v dv = ln|v| + C

8. ∫ 1/cos(u) du = ln|cos(u)| + C

9. ∫ sec^2(u) du = ln|cos(u)| + C

10. ∫ sec^2(7−4x) du = ln|cos(7−4x)| + C

Question:

Integrate the function xlogx

Answer:

1. Rewrite the function as log(x^x)
2. Use the integration by parts technique:

Let u = log(x^x) and dv = dx

Then du = (xlnx + 1)dx and v = x

Integrating both sides, we get:

∫u dv = uv - ∫v du

3. Substitute the values of u, dv, v and du:

∫log(x^x)dx = xlog(x^x) - ∫x(xlnx + 1)dx

4. Integrate the right hand side:

∫log(x^x)dx = xlog(x^x) - (x^2lnx + x^2)/2 + C

5. Simplify the result:

∫log(x^x)dx = (x^2lnx + x^2 - 2xlog(x^x))/2 + C

Question:

Integrate the function 1/cos^2x(1−tanx)^2

Answer:

1. Integrate (1/cos^2x) = sinx/cosx

2. Integrate (1−tanx)^2 = (1−tanx)(1−tanx) = 1−2tanx+tan^2x

3. Multiply the two integrals = sinx/cosx - 2sinx/cosx + sinx/cosx tan^2x = sinx/cosx (1 - tan^2x)

4. Integrate (1 - tan^2x) = x - tanx

5. Substitute the two integrals = (sinx/cosx)(x - tanx) = sinx - sinxtanx

Question:

Integrate the function √sin2xcos2x

Answer:

Answer: Step 1: Rewrite the function as sin^2xcos^2x Step 2: Use the formula ∫sin^2xcos^2xdx = (1/2)∫(1-cos2x)dx Step 3: Integrate (1-cos2x) using the formula ∫(1-cos2x)dx = x - (1/2)sin2x Step 4: Substitute the result from Step 2 into the original equation Step 5: Final answer: (1/2)x - (1/4)sin2x

Question:

Integrate the function e^x(1​/x−1​/x^2)

Answer:

Step 1: Apply the power rule for integration.

Integral of e^x(1/x - 1/x^2)dx = e^x(ln|x| - 1/x) + C

Question:

∫dx​/√9x−4x^2 equals A 1​/9sin^−1(9x−8/8​)+C B 1​/2sin^−1(8x−9/9​)+C C 1​/3sin^−1(9x−8/8​)+C D 1​/2sin^−1(9x−8/9​)+C

Answer:

Answer: C

Step 1: Rewrite the integral as ∫dx/√(9x - 4x²).

Step 2: Complete the square to get ∫dx/√(9x - 4x²) = ∫dx/√(9(x - 8/9))

Step 3: Substitute u = 9x - 8 to get ∫dx/√(9x - 4x²) = 1/9∫du/√u.

Step 4: Use the substitution rule to get ∫dx/√(9x - 4x²) = 1/9∫du/√u = 1/9sin^−1(u/9)+C.

Step 5: Substitute u = 9x - 8 to get ∫dx/√(9x - 4x²) = 1/9sin^−1(9x - 8/9)+C.

Therefore, the answer is C.

Question:

Integrate the function 1/√7−6x−x^2

Answer:

1. Rewrite the function in standard form: f(x) = 1/√7 - 6x - x^2

2. Integrate with respect to x: ∫f(x)dx = ∫(1/√7 - 6x - x^2)dx

3. Use the power rule: ∫x^n dx = (x^(n+1))/(n+1)

4. Apply the power rule: ∫(1/√7 - 6x - x^2)dx = (1/√7x - 3x^2 - x^3/3) + C

5. The answer is: (1/√7x - 3x^2 - x^3/3) + C

Question:

Integrate the function 1/√(2−x)^2+1

Answer:

Answer: Step 1: Rewrite the function as 1/[(√2 - x)(√2 + x)] + 1

Step 2: Rewrite the denominator as a perfect square, (√2 - x)^2 + (√2 + x)^2

Step 3: Rewrite the function as 1/[(√2 - x)^2 + (√2 + x)^2] + 1

Step 4: Take the integral of the function, ∫1/[(√2 - x)^2 + (√2 + x)^2] + 1dx

Step 5: Use the substitution u = √2 - x and du = -dx

Step 6: Substitute u and du into the integral and solve, -1/2 ∫(1/u^2 + 1)du

Step 7: Integrate the function, -1/2(u^-1 + u) + C

Step 8: Substitute back in x and solve, -1/2(√2 - x)^-1 + (√2 - x) + C

Question:

Find an anti derivative (or integral) of the given function by the method of inspection. sin2x−4e^3x

Answer:

Answer:

1. Rewrite the given function using the sum and difference rules: sin2x−4e^3x = 2sinxcosx−4e^3x
2. Use the product rule to break down the function: 2sinxcosx−4e^3x = 2sinx(cosx−2e^3x)
3. Integrate both terms separately: ∫2sinx(cosx−2e^3x)dx = 2∫sinxcosxdx−2∫e^3xdx
4. Use the integration by parts formula to integrate the first term: 2∫sinxcosxdx = sin2x−2∫sinxsinxdx
5. Use the substitution rule to integrate the second term: 2∫e^3xdx = 2e^3x
6. Combine the two terms to get the anti derivative: sin2x−2∫sinxsinxdx−2e^3x

Question:

Integrate the function sin(ax+b)cos(ax+b)

Answer:

1. Use the formula ∫sin(u)cos(v)dv = 1/2(sin(u+v) - sin(u-v))

2. Substitute u = ax + b and v = ax + b:

∫sin(ax+b)cos(ax+b)dv = 1/2(sin(2ax + 2b) - sin(0))

3. Simplify:

∫sin(ax+b)cos(ax+b)dv = 1/2(sin(2ax + 2b))

Question:

∫(sin^2−cos^2x​/sin^2xcos^2x)dx is equal to A tanx+cotx+C B tanx+cosecx+C C −tanx+cotx+C D tanx+secx+C

Answer:

Answer: C

Step-by-step solution:

Step 1: Rewrite the given integral as

∫(sin^2x - cos^2x)/(sin^2xcos^2x)dx

Step 2: Use the trigonometric identity sin2x = 2sinxcosx to rewrite the integral as

∫(2sinxcosx - cos^2x)/(sin^2xcos^2x)dx

Step 3: Use the trigonometric identity cos2x = cos^2x - sin^2x to rewrite the integral as

∫(-sin^2x)/(sin^2xcos^2x)dx

Step 4: Use the trigonometric identity 1/sin2x = cot2x to rewrite the integral as

∫(-cot2x)dx

Step 5: Integrate the expression to get

-tanx + cotx + C

Question:

​Find the integral of ∫(4e^3x+1)dx

Answer:

Answer:

Step 1: Rewrite the integral as: ∫4e^3x dx + ∫1 dx

Step 2: Integrate the first term: ∫4e^3x dx = (1/3)e^3x + C

Step 3: Integrate the second term: ∫1 dx = x + C

Step 4: Combine the two results: (1/3)e^3x + x + C

Question:

Integrate the function x−1/√x^2−1

Answer:

Answer: Step 1: Rewrite the function as (x-1)/(x+1)(x-1) Step 2: Factor the denominator to get (x-1)(x+1)/(x+1)(x-1) Step 3: Use the substitution u = x+1 Step 4: Rewrite the function as u-2/u^2 Step 5: Integrate the function to get 1/2u + C, where C is a constant. Step 6: Substitute x+1 for u to get 1/2(x+1) + C

Question:

Find the integrals of the functions sinxsin2xsin3x

Answer:

Answer:

1. Integrate sin x: ∫sinx dx = -cosx + C
2. Integrate sin2x: ∫sin2x dx = -1/2 cos2x + C
3. Integrate sin3x: ∫sin3x dx = -1/3 cos3x + C

Therefore, the integral of sinxsin2xsin3x is ∫sinxsin2xsin3x dx = (-cosx)(-1/2 cos2x)(-1/3 cos3x) + C = 1/6 cosxcos2xcos3x + C

Question:

Find the integrals of the functions sin4xsin8x

Answer:

1. Use the identity sinA sinB = (1/2) [cos(A-B) - cos(A+B)]

2. Integrate both sides with respect to x

3. Integrate the left side using the standard formula for the integral of a sine function: ∫sinA dx = -cosA + C

4. Integrate the right side using the standard formula for the integral of a cosine function: ∫cosA dx = sinA + C

5. Substitute the identity into the integrals and simplify

6. Combine the constants of integration and simplify -cos(4x)sin(8x) + sin(4x)cos(8x) + C

Question:

If the integrals of the function sin^2x​/1+cosx is ax+bsinx.Find a+b.

Answer:

Step 1: Integrate sin2x / 1 + cosx Step 2: a sinx + b cosx + c Step 3: Set c = 0 Step 4: Compare the coefficients of sinx and cosx Step 5: a + b = 0 Step 6: Thus, a + b = 0

Question:

Find the integral of (x^3+3x+4​/∫x)dx

Answer:

1. Rewrite the integral as ∫(x^3+3x+4)dx

2. Use the power rule to integrate ∫x^3dx + ∫3xdx + ∫4dx

3. Solve each integral 1/4x^4 + x^2 + 4x + C

4. The final answer is 1/4x^4 + x^2 + 4x + C

Question:

Integrate the function xcos^−1x

Answer:

1. Rewrite the function as cos^−1x/x
2. Use integration by parts with u = cos^−1x and dv = 1/x
3. du = -1/(x√(1-x2)) and v = ln|x|
4. Integrate: cos^−1xln|x| - ∫1/x(-(1/(x√(1-x2))))dx
5. Use integration by parts again with u = 1/x and dv = -1/(x√(1-x2))
6. du = -1/(x2) and v = -(1/2)ln(1-x2)
7. Integrate: cos^−1xln|x| - (1/2)ln(1-x2) + ∫(1/(x2))(1/(2√(1-x2)))dx
8. Integrate: cos^−1xln|x| - (1/2)ln(1-x2) + (1/2)√(1-x2) + C

Question:

Integrate the function √1−4x−x^2

Answer:

1. Rewrite the function as (1 - x)^(1/2) - x^2

2. Integrate (1 - x)^(1/2) using u-substitution: Let u = 1 - x du = -dx

3. Integrate -x^2 using the power rule: -x^2 –> -1/3x^3

4. Substitute u back in: -1/3(1 - x)^3

5. The final answer is -1/3(1 - x)^3

Question:

Integrate the rational function x^3+x+1/x^2−1

Answer:

1. Rewrite the rational function as (x^3 + x + 1)(x^2 + 1)^-1

2. Use the formula for integrating a rational function: Integral (f/g)dx = Integral fdx/g - Integral (f’g - fg’)/g^2 dx

3. Integrate fdx/g Integral (x^3 + x + 1)dx/ (x^2 + 1) = (x^4/4 + x^2/2 + x) / (x^2 + 1)

4. Integrate (f’g - fg’)/g^2 dx Integral (3x^2 - (x^3 + x + 1))(x^2 + 1)^-2 dx = -(x^4/4 + x^2/2 + x) / (x^2 + 1)

5. Add the two integrals together: Integral (x^3 + x + 1)dx/ (x^2 + 1) + Integral (3x^2 - (x^3 + x + 1))(x^2 + 1)^-2 dx = 0

Question:

Find an anti derivative (or integral) of the given function by the method of inspection. (ax+b)^2

Answer:

Answer: Step 1: Recognize that the given function is in the form (ax+b)^2, which can be written as a^2x^2 + 2abx + b^2.

Step 2: Apply the method of inspection to determine the anti derivative. The anti derivative of a^2x^2 is (1/3)a^2x^3, the anti derivative of 2abx is (1/2)abx^2, and the anti derivative of b^2 is b^2x.

Step 3: Combine the terms to obtain the anti derivative. The anti derivative of (ax+b)^2 is (1/3)a^2x^3 + (1/2)abx^2 + b^2x.

Question:

Integrate the function e^x(1+sinx​/1+cosx)

Answer:

Answer: Step 1: Rewrite the function in terms of sin2x and cos2x e^x(1+sinx)/(1+cosx) = e^x(1+sin2x/2)/(1+cos2x/2)

Step 2: Use the substitution u = sin2x e^x(1+sin2x/2)/(1+cos2x/2) = e^x(1+u/2)/(1+√(1-u^2)/2)

Step 3: Integrate ∫e^x(1+u/2)/(1+√(1-u^2)/2)du = 2/3e^x(u+2√(1-u^2)-3ln(1+√(1-u^2))+C

Question:

Find an anti derivative (or integral) of the given function by the method of inspection. e^2x

Answer:

1. First, determine if the function is in the form of a polynomial. Since e^2x is an exponential function, it is not in the form of a polynomial.

2. Therefore, use the method of inspection to find the anti derivative of e^2x.

3. To do this, take the natural logarithm of both sides of the equation.

4. This gives us: ln(e^2x) = ln(y).

5. Then, use the power rule for logarithms to solve for y: y = 2x*e^2x.

6. Finally, the anti derivative of e^2x is 2x*e^2x.

Question:

∫e^x(1+x)​/cos^2(e^xx)dx equals A −cot(ex^x)+C B tan(xe^x)+C C tan(e^x)+C D cot(e^x)+C

Answer:

Answer: B tan(xe^x)+C

Question:

Find the integral of the function cos2x/(cosx+sinx)^2

Answer:

Answer: Step 1: Rewrite the function as (1-sinx^2)/(cosx+sinx)^2

Step 2: Take the derivative of the denominator: (cosx+sinx)^2 = 2cosxsinx

Step 3: Rewrite the function as (1-sinx^2)/2cosxsinx

Step 4: Take the derivative of the numerator: -2sinx

Step 5: Rewrite the function as -sinx/2cosxsinx

Step 6: Integrate with respect to x: -1/2ln(cosx+sinx)+C

Question:

Integrate the function sinx/(1+cosx)^2

Answer:

Step 1: Rewrite the function using the identity sinx = (1+cosx) - (1-cosx) sinx/(1+cosx)^2 = (1+cosx - (1-cosx))/(1+cosx)^2

Step 2: Rewrite the denominator using the identity (1+cosx)^2 = 1 + 2cosx + cos^2x sinx/(1+cosx)^2 = (1+cosx - (1-cosx))/(1 + 2cosx + cos^2x)

Step 3: Rewrite the numerator using the identity 1+cosx = 2cos^2x/2 sinx/(1+cosx)^2 = (2cos^2x/2 - (1-cosx))/(1 + 2cosx + cos^2x)

Step 4: Factor out the common term 2cos^2x/2 from the numerator sinx/(1+cosx)^2 = (2cos^2x/2)(1/(1 + 2cosx + cos^2x) - (1-cosx)/(2cos^2x/2))

Step 5: Simplify the expression sinx/(1+cosx)^2 = (1/(1 + 2cosx + cos^2x) - (1-cosx)/(2cos^2x))

Step 6: Integrate both sides ∫sinx/(1+cosx)^2dx = ∫(1/(1 + 2cosx + cos^2x) - (1-cosx)/(2cos^2x))dx

Step 7: Solve the integrals on the right side ∫sinx/(1+cosx)^2dx = (1/2)ln|1 + 2cosx + cos^2x| - (1/2)ln|2cos^2x| - tanx + C

Question:

Integrate the rational function x/(x−1)^2(x+2)

Answer:

1. Integrate the rational function (1/x) * (1/(x-1)^2) * (1/(x+2))

2. Integrate (1/x) ln|x| + C

3. Integrate (1/(x-1)^2) 1/(x-1) + C

4. Integrate (1/(x+2)) ln|x+2| + C

5. Combine the answers ln|x| + 1/(x-1) + ln|x+2| + C

Question:

Integrate the function x^3e^x

Answer:

Answer:

1. Take the antiderivative of x^3: ∫x^3dx = (1/4)x^4 + C

2. Take the antiderivative of e^x: ∫e^xdx = e^x + C

3. Multiply the two antiderivatives together: (1/4)x^4e^x + C

4. Take the antiderivative of the product: ∫(1/4)x^4e^xdx = (1/4)x^4e^x + C

Question:

ntegrate the function √1+x^2/9

Answer:

Answer: Step 1: Integrate (1+x^2)^(1/2) with respect to x.

Step 2: Use the power rule to integrate.

Step 3: (1+x^2)^(1/2) = u, u = (1+x^2)^(1/2)

Step 4: du/dx = (1/2)(1+x^2)^(-1/2)(2x)

Step 5: Integrate (1+x^2)^(1/2) with respect to x

Step 6: ∫(1+x^2)^(1/2) dx = ∫u du

Step 7: = (1/2)u^2 + c

Step 8: = (1/2)(1+x^2) + c

Step 9: Integrate (√1+x^2)/9 with respect to x

Step 10: Divide (1+x^2)^(1/2) by 9

Step 11: = (1/9)(1+x^2)^(1/2)

Step 12: Integrate (1/9)(1+x^2)^(1/2) with respect to x

Step 13: = (1/9)(1/2)(1+x^2) + c

Step 14: = (1/18)(1+x^2) + c

Hence, the answer is ∫ (√1+x^2)/9 dx = (1/18)(1+x^2) + c

Question:

Integrate the function 1/√(x−a)(x−b)

Answer:

1. Separate the function into two parts: 1/√(x-a) and 1/(x-b)

2. Integrate each part separately: ∫1/√(x-a)dx = 2√(x-a) + C ∫1/(x-b)dx = ln|x-b| + C

3. Combine the two parts: 2√(x-a) + ln|x-b| + C

Question:

Integrate the function 1/√9−25x^2

Answer:

Answer: Step 1: Rewrite the function as: 1/√(9 - 25x^2)

Step 2: Use the substitution u = 9 - 25x^2

Step 3: Rewrite the function as: 1/√u

Step 4: Integrate the function using the following formula: 2/3u^(3/2) + C

Question:

​Integrate the function xe^x/(1+x)^2

Answer:

Answer:

Step 1: Rewrite the function as xe^x/[(1+x) (1+x)]

Step 2: Integrate xe^x with respect to x: ∫xe^x dx = e^x(x + 1)

Step 3: Integrate 1/(1+x) with respect to x: ∫1/(1+x) dx = ln(1+x)

Step 4: Integrate 1/(1+x)^2 with respect to x: ∫1/(1+x)^2 dx = 1/[(1+x)(ln(1+x))]

Step 5: Multiply the results: e^x(x + 1) x 1/[(1+x)(ln(1+x))]

Step 6: Simplify the expression: e^x(x + 1) / ln(1+x)

Question:

Integrate the function cos√x/√x

Answer:

1. u = √x
2. du = (1/2)x^(-1/2)dx
3. Integral of cosu*du = sinu + C
4. Integral of cos√x/√x = sin√x + C

Question:

​∫x^2e^x^3 dx equals A 1/3​e^x^3+C B 1/3​e^x^2+C C 1/2​e^x^3+C D e^−x^2+C

Answer:

Answer: C

Solution:

Step 1: Use the integration by parts formula.

Integration by parts formula: ∫uvdx = uv - ∫vdu

Let u = x^2 and dv = e^x^3 dx

Step 2: Calculate du and v.

du = 2x dx

v = 1/3e^x^3

Step 3: Substitute the values of u, du, and v in the integration by parts formula.

∫x^2e^x^3 dx = x^2*(1/3e^x^3) - ∫(2x)*(1/3e^x^3) dx

Step 4: Integrate the second term.

∫(2x)*(1/3e^x^3) dx = 2/3e^x^3 + C

Step 5: Substitute the value of the second term in the first equation.

∫x^2e^x^3 dx = x^2*(1/3e^x^3) - (2/3e^x^3 + C)

Step 6: Simplify the equation.

∫x^2e^x^3 dx = 1/3e^x^3 + C

Question:

Integrate the rational function 5x/(x+1)(x^2−4)

Answer:

Step 1: Multiply 5x by the conjugate of the denominator, (x−2)(x+2).

Step 2: Simplify the denominator, (x+1)(x−2)(x+2) = x^3 + x^2 - 4x - 8.

Step 3: Use the method of partial fractions to decompose the rational expression.

Step 4: Integrate each term in the decomposition separately.

Step 5: Add the resulting integrals together to obtain the final answer.

Answer: 5/2ln|x+2| - 5/2ln|x-2| - 5/3x^3 + 5/2x^2 + 10x + C

Question:

Integrate the function e^2xsinx

Answer:

1. ∫e^2xsinx dx

2. ∫e^2x sinx dx

3. u = e^2x, du = 2e^2x dx

4. dv = sinx dx, v = -cosx

5. ∫e^2x sinx dx = -e^2x cosx + ∫2e^2x cosx dx

6. u = e^2x, du = 2e^2x dx

7. dv = cosx dx, v = sinx

8. ∫e^2x sinx dx = -e^2x cosx + 2e^2x sinx + C

Question:

Integrate the function xsec^2x

Answer:

Answer:

1. ∫xsec^2x dx
2. ∫(sec^2x)dx
3. ∫(1/cos^2x)dx
4. ∫(tan^2x + 1)dx
5. ∫(sec^2x tan^2x + sec^2x)dx
6. ∫sec^3x tan^2x + ∫secx dx
7. 1/2 sec^2x tanx + tanx + C

Question:

Find the integral of ∫secx(secx+tanx)dx

Answer:

1. ∫secx(secx+tanx)dx

2. ∫secxsecxdx + ∫secxtandx

3. ∫sec²xdx + ∫secxtandx

4. ∫sec²xdx + ∫sec²xdx - ∫sec²xdx

5. ∫sec²xdx + ∫sec²xdx - ∫tan²xdx

6. tanx + secx - ∫tan²xdx

7. tanx + secx - ∫sec²tanx dx

8. tanx + secx - ∫secx dx + ∫tanx dx

9. tanx + secx - ln|secx + tanx| + tanx + C

Question:

Find the integral of ∫(ax^2+bx+c)dx

Answer:

Answer: Step 1: Rewrite the equation as ∫ax^2 + bx + c dx.

Step 2: Integrate ax^2.

∫ax^2 dx = (1/3)ax^3 + C

Step 3: Integrate bx.

∫bx dx = (1/2)bx^2 + C

Step 4: Integrate c.

∫c dx = cx + C

Step 5: Combine all the terms.

∫(ax^2 + bx + c)dx = (1/3)ax^3 + (1/2)bx^2 + cx + C

Question:

Find the integrals of the functions cosx/1+cosx

Answer:

1. Rewrite the function as cosx/(1+cosx) = cosx * 1/(1+cosx)

2. Use the substitution u = 1 + cosx

3. du = -sinx dx

4. Integrate: ∫cosx/(1+cosx)dx = ∫cosx * 1/(1+cosx)dx = ∫cosx * 1/u du = ∫-cosx/u du

5. Integrate: -∫cosx/u du = -sinx/u + C

6. Substitute back u = 1 + cosx

7. -sinx/(1+cosx) + C

Question:

Find the integral of ∫(2x−3cosx+e^x)dx

Answer:

1. ∫(2x−3cosx+e^x)dx

2. ∫2xdx − 3∫cosxdx + ∫e^xdx

3. (1/2)x^2 - 3sinx + e^x + C

Question:

Find the integrals of the function cos2x−cos2α/cosx−cosα

Answer:

Answer:

Step 1: Rewrite the given expression using the identity cos2x = 1 - 2sin2x

cos2x − cos2α/cosx − cosα = (1 - 2sin2x - cos2α)/(cosx − cosα)

Step 2: Rewrite the denominator using the identity cos2α = 1 - 2sin2α

cos2x − cos2α/cosx − cosα = (1 - 2sin2x - (1 - 2sin2α))/(cosx − (1 - 2sin2α))

Step 3: Multiply both numerator and denominator by (cosx + (1 - 2sin2α))

cos2x − cos2α/cosx − cosα = (cosx + (1 - 2sin2α))(1 - 2sin2x - (1 - 2sin2α))/(cos2x - 2cosxsin2α + (1 - 2sin2α)^2)

Step 4: Integrate both sides with respect to x

∫cos2x − cos2α/cosx − cosα dx = ∫(cosx + (1 - 2sin2α))(1 - 2sin2x - (1 - 2sin2α))/(cos2x - 2cosxsin2α + (1 - 2sin2α)^2) dx

Step 5: Solve the integral

∫cos2x − cos2α/cosx − cosα dx = (1/2)ln|cos2x - 2cosxsin2α + (1 - 2sin2α)^2| + C

Question:

Find the integrals of the functions sin2xcos2xsin3x+cos3x

Answer:

Answer:

1. Integrate sin2xcos2x: ∫sin2xcos2xdx = -1/2cos4x + C

2. Integrate sin3x: ∫sin3xdx = -1/2cos3x + C

3. Integrate cos3x: ∫cos3xdx = 1/2sin3x + C

4. Add the results from steps 1-3: -1/2cos4x - 1/2cos3x + 1/2sin3x + C

Question:

Integrate the function 3x^2/x^6+1

Answer:

Step 1: Rewrite the function as 3x^2/(x^6+1).

Step 2: Use the substitution u = x^6 + 1.

Step 3: Rewrite the function as 3x^2/u.

Step 4: Take the derivative of u with respect to x: du/dx = 6x^5.

Step 5: Multiply the function by du/dx: 3x^2*6x^5/u.

Step 6: Rewrite the function as 18x^7/u.

Step 7: Integrate the function with respect to x: ∫18x^7/u dx.

Step 8: Substitute u = x^6 + 1: ∫18x^7/(x^6+1) dx.

Step 9: Integrate the function: 18/6 (x^6+1)^(1/6) + C, where C is the constant of integration.

Question:

Find the integral of ∫(2x^2−3sinx+5√x)dx

Answer:

1. Rewrite the integrand using the power rule:

∫(2x^3 - 3cosx + 5x^(1/2))dx

2. Integrate:

2/4 x^4 - 3/2 sinx + 5/3 x^(3/2) + C

Question:

Find the integral of the function 1/cos(x−a)cos(x−b)

Answer:

1. Use the substitution rule by setting u = cos(x−a):

∫1/cos(x−a)cos(x−b)dx = ∫1/u(cos(x−b))dx

2. Use the product rule to break up the integral:

∫1/u(cos(x−b))dx = ∫1/u du + ∫cos(x−b)dx

3. Integrate the first term:

∫1/u du = ln|u| + C

4. Integrate the second term:

∫cos(x−b)dx = sin(x−b) + C

5. Substitute back in the original variable and constants:

ln|cos(x−a)| + sin(x−b) + C

Question:

Integrate the function (1+logx)^2/x

Answer:

1. Rewrite the function as: (1+lnx)^2/x

2. Use the power rule to integrate: (1+lnx)^2/x = (1+lnx)^2/x * (1/1+lnx)

3. Rewrite the function as: (1+lnx)^3/1+lnx

4. Use the power rule to integrate: (1+lnx)^3/1+lnx = (1+lnx)^3/1+lnx * (1/(3*(1+lnx)^2))

5. Rewrite the function as: (1+lnx)^3/(3*(1+lnx)^2)

6. Use the power rule to integrate: (1+lnx)^3/(3*(1+lnx)^2) = (1+lnx)^3/(3*(1+lnx)^2) * (1/(3*(1+lnx)))

7. Rewrite the function as: (1+lnx)^4/(9*(1+lnx)^3)

8. Use the power rule to integrate: (1+lnx)^4/(9*(1+lnx)^3) = (1+lnx)^4/(9*(1+lnx)^3) * (1/(4*(1+lnx)^2))

9. Rewrite the function as: (1+lnx)^5/(36*(1+lnx)^4)

10. Use the power rule to integrate: (1+lnx)^5/(36*(1+lnx)^4) = (1+lnx)^5/(36*(1+lnx)^4) * (1/(5*(1+lnx)^3))

11. Rewrite the function as: (1+lnx)^6/(216*(1+lnx)^5)

12. Use the power rule to integrate: (1+lnx)^6/(216*(1+lnx)^5) = (1+lnx)^6/(216*(1+lnx)^5) * (1/(6*(1+lnx)^4))

13. Rewrite the function as: (1+lnx)^7/(1296*(1+lnx)^6)

14. Use the power rule to integrate: (1+lnx)^7/(1296*(1+lnx)^6) = (1+lnx)^7/(1296*(1+lnx)^6) * (1/(7*(1+lnx)^5))

15. Rewrite the function as: (1+lnx)^8/(7776*(1+lnx)^7)

16. Use the power rule to integrate: (1+lnx)^8/(7776*(1+lnx)^7) = (1+lnx)^8/(7776*(1+lnx)^7) * (1/(8*(1+lnx)^6))

17. Rewrite the function as: (1+lnx)^9/(46656*(1+lnx)^8)

18. Use the power rule to integrate: (1+lnx)^9/(46656*(1+lnx)^8) = (1+lnx)^9/(46656*(1+lnx)^8) * (1/(9*(1+lnx)^7))

19. Rewrite the function as: (1+lnx)^10/(279936*(1+lnx)^9)

20. Use the power rule to integrate: (1+lnx)^10/(279936*(1+lnx)^9) = (1+lnx)^10/(279936*(1+lnx)^9) * (1/(10*(1+lnx)^8))

21. The final answer is: (1+lnx)^11/(2799360*(1+lnx)^10)

Question:

Integrate the function x/e^x^2

Answer:

Step 1: Rewrite the function as xe^(-x^2)

Step 2: Integrate both sides with respect to x:

Step 3: ∫xe^(-x^2)dx

Step 4: Use the integration by parts formula:

Step 5: u = x and dv = e^(-x^2)dx

Step 6: du = dx and v = -1/2e^(-x^2)

Step 7: ∫xe^(-x^2)dx = x(-1/2e^(-x^2)) - ∫(-1/2e^(-x^2))dx

Step 8: Integrate the second term on the right-hand side:

Step 9: ∫(-1/2e^(-x^2))dx = -1/2∫e^(-x^2)dx

Step 10: Use the substitution u = x^2:

Step 11: du = 2xdx and ∫e^(-x^2)dx = 1/2∫e^(-u)du

Step 12: ∫e^(-u)du = -e^(-u)

Step 13: Substitute u = x^2:

Step 14: -e^(-u) = -e^(-x^2)

Step 15: Final answer:

∫xe^(-x^2)dx = x(-1/2e^(-x^2)) + 1/2e^(-x^2)

Question:

Integrate the function 1/1−tanx

Answer:

1. Use the substitution u = tanx
2. du = sec^2x dx
3. ∫1/(1-tanx)dx = ∫1/u du
4. ∫1/u du = ln|u| + C
5. ∫1/(1-tanx)dx = ln|tanx| + C

Question:

Integrate the function sin^−1x/√1−x^2

Answer:

1. Let u = sin^−1x
2. du = 1/√1−x^2 dx
3. Integrate u du
4. ∫udu = u^2/2 + C
5. ∫sin^−1x/√1−x^2dx = (sin^−1x)^2/2 + C

Question:

Integrate the function (x+1)(x+logx)^2/x

Answer:

Answer: Step 1: Use the product rule to expand the given function: (x + 1)(x + logx)2/x = (x + logx)2 + 2(x + logx) + 1

Step 2: Integrate each term separately: (x + logx)2 = (1/3)(x + logx)3 + C 2(x + logx) = (2/3)(x + logx)3 + C 1 = x + C

Step 3: Add all the terms together to get the final solution: Integral of (x + 1)(x + logx)2/x = (1/3)(x + logx)3 + (2/3)(x + logx)3 + x + C

Question:

Integrate the function tan^−1x

Answer:

1. Rewrite the function as arctan(x)
2. Integrate arctan(x) with respect to x
3. ∫arctan(x)dx = xarctan(x) - ∫1/(1+x^2)dx
4. ∫1/(1+x^2)dx = ∫1/u du (where u = 1+x^2)
5. ∫1/u du = ln|u| + c
6. ∫arctan(x)dx = xarctan(x) - ln|1+x^2| + c

Question:

Integrate the function xcos^−1x/√1−x^2

Answer:

1. Let u = cos^−1x
2. du = -1/√1-x^2 dx
3. ∫ xcos^−1x/√1−x^2 dx
4. ∫ xdu/√1−x^2
5. ∫ u/√1−x^2 du
6. 1/2∫ (u√1−x^2 + x^2√1−x^2) du
7. 1/2[u√1−x^2 + x^2√1−x^2] + C

Question:

Integrate the function 3x/1+2x^4

Answer:

1. Rewrite the function as 3x(1+2x^4)^-1

2. Use the formula ∫u^-1du = ln|u| + C

3. Integrate the function: ∫3x(1+2x^4)^-1 dx = ln|1+2x^4| + C

Question:

Integrate the function ​1/√8+3x−x^2

Answer:

1. Separate the function into two parts: 1/√8 + 3x and -x2
2. Integrate the first part: 2/3√8x + C
3. Integrate the second part: -1/3x3 + C
4. Combine the two parts together: 2/3√8x - 1/3x3 + C

Question:

Integrate the function √1+3x−x^2

Answer:

1. ∫√1+3x−x^2dx
2. ∫(1+3x−x^2)^(1/2)dx
3. Let u = 1+3x−x^2
4. du = 3−2x dx
5. ∫u^(1/2)du
6. 2/3u^(3/2) + C
7. 2/3(1+3x−x^2)^(3/2) + C

Question:

Integrate the function √x^2+3x

Answer:

1. Rewrite the function as x^2 + 3x = (x + 3)^2 - 9

2. Integrate (x + 3)^2 - 9 dx

3. Integrate (x + 3)^2 dx

4. ∫(x + 3)^2 dx = (1/3)(x + 3)^3 + C

Question:

Integrate the function x^2logx

Answer:

Step 1: Rewrite the function in a form that can be integrated: x^2logx = x^2lnx

Step 2: Integrate the function: ∫x^2lnx dx = x^3(lnx - 1) + C

Question:

Integrate the function xlog2x

Answer:

Step 1: Rewrite the function as log2x + x*(d/dx of log2x)

Step 2: Integrate both sides of the equation with respect to x

Step 3: Calculate the integral of log2x

Integral of log2x = x*log2x - (1/ln(2))*x2

Step 4: Calculate the integral of (d/dx of log2x)

Integral of (d/dx of log2x) = x

Step 5: Substitute the integrals into the equation

x*log2x - (1/ln(2))*x2 + x = Integral

Step 6: Simplify the equation

Integral = x*log2x - (1/ln(2))*x2 + x

Question:

Integrate the function (sin^−1x)^2

Answer:

1. Rewrite the function as (sin^-1x)^2 = (sin^2x)^-1

2. Take the inverse of both sides: (sin^-1x)^2 = (sin^2x)^-1 → (sin^2x) = (sin^-1x)^2

3. Take the derivative of both sides with respect to x: (sin^2x)’ = (sin^-1x)^2'

4. Simplify the derivative on the left side: (sin^2x)’ = 2sinxcosx

5. Simplify the derivative on the right side: (sin^-1x)^2’ = 2(sin^-1x)(cos^-1x)

6. Set the derivatives equal to each other and solve for cos^-1x: 2sinxcosx = 2(sin^-1x)(cos^-1x) → cos^-1x = (sinx)/(sin^-1x)

7. Integrate both sides with respect to x: ∫cos^-1x dx = ∫(sinx)/(sin^-1x) dx

8. Use the substitution u = sin^-1x: ∫cos^-1x dx = ∫du/u dx

9. Integrate both sides: ∫cos^-1x dx = ln|u| + C

10. Substitute back in u = sin^-1x: ∫cos^-1x dx = ln|sin^-1x| + C

Question:

Integrate the rational function: 1/x^2−9

Answer:

Answer: Step 1: Separate the fraction into two separate fractions: 1/x^2 and -9. Step 2: Integrate each fraction separately. Step 3: For the first fraction, use the power rule to integrate: ∫1/x^2dx = -1/x + C Step 4: For the second fraction, use the constant rule to integrate: ∫-9dx = -9x + C Step 5: Combine the two results to get the final answer: -1/x - 9x + C

Question:

Integrate the rational function x/(x+1)(x+2)

Answer:

Step 1: Use the method of partial fraction decomposition to split the rational function into simpler fractions.

x/(x+1)(x+2) = A/(x+1) + B/(x+2)

Step 2: Solve for the constants A and B by multiplying both sides by (x+1)(x+2).

A(x+2) + B(x+1) = x

Step 3: Solve for A by setting x = -2 and substituting into the equation.

A(-2+2) + B(-2+1) = -2 A(0) + B(-1) = -2 A = 2

Step 4: Solve for B by setting x = -1 and substituting into the equation.

2(-1+2) + B(-1+1) = -1 2(-1) + B(0) = -1 B = 1

Step 5: Substitute the values of A and B into the partial fraction decomposition.

x/(x+1)(x+2) = 2/(x+1) + 1/(x+2)

Step 6: Integrate both sides of the equation.

∫x/(x+1)(x+2)dx = ∫2/(x+1)dx + ∫1/(x+2)dx

Step 7: Solve each integral.

∫2/(x+1)dx = 2 ln|x+1| + C1 ∫1/(x+2)dx = ln|x+2| + C2

Step 8: Combine the two integrals.

∫x/(x+1)(x+2)dx = 2 ln|x+1| + ln|x+2| + C

Step 9: Simplify the equation.

∫x/(x+1)(x+2)dx = ln|(x+1)(x+2)| + C

Question:

Integrate the rational function 2x−3/(x^2−1)(2x+3)

Answer:

1. Rewrite the numerator and denominator as the product of two linear factors: Numerator: 2x−3 = (2x−3)(1) Denominator: x^2−1 = (x−1)(x+1)

2. Rewrite the rational function as the sum of two fractions: 2x−3/(x^2−1)(2x+3) = (2x−3)(1)/((x−1)(x+1))(2x+3) + (2x−3)(2x+3)/((x−1)(x+1))(2x+3)

3. Integrate each fraction separately: (2x−3)(1)/((x−1)(x+1))(2x+3) = 1/2 ln|x+1| - 1/2 ln|x-1| (2x−3)(2x+3)/((x−1)(x+1))(2x+3) = x + 3/2 ln|x+1| - 3/2 ln|x-1|

4. Add the two results together to get the full answer: Integral of 2x−3/(x^2−1)(2x+3) = x + 1/2 ln|x+1| - 1/2 ln|x-1|

Question:

Integrate the function √1−4x^2

Answer:

Answer: Step 1: Rewrite the function as 1 - 4x^2 = (1 - 2x)(1 + 2x)

Step 2: Use the formula for integration of a product of two terms: ∫(f(x)g(x))dx = ∫f(x)dx∫g(x)dx

Step 3: Integrate each term separately: ∫(1 - 2x)dx = x - x^2 + C ∫(1 + 2x)dx = x + x^2 + C

Step 4: Combine the two terms: x - x^2 + C + x + x^2 + C = 2x + 2C

Question:

Integrate the function x​/√x+4,x>0

Answer:

Answer: Step 1: Rewrite the integrand as a fraction: x/(x+4)1/2

Step 2: Use the power rule to integrate: (2/3)x3/2/(x+4)1/2 + C

Step 3: Simplify the result: (2/3)(x3/2 + 4x1/2) + C

Question:

Integrate the function tan^2(2x−3)

Answer:

1. Use the formula for integration of tan^2(u): ∫tan^2(u)du = (1/2)tan(u) - (1/2)ln|sec(u)+tan(u)| + C

2. Substitute 2x−3 for u: ∫tan^2(2x−3)dx = (1/2)tan(2x−3) - (1/2)ln|sec(2x−3)+tan(2x−3)| + C

3. Simplify the equation: ∫tan^2(2x−3)dx = (1/2)tan(2x−3) - (1/2)ln|sec(2x−3)+tan(2x−3)| + C

Question:

Integrate the function (x^3−1)^1​/3x^5

Answer:

1. Rewrite the function using the power rule: x^3−1 = (x^2)^1/2⋅x−1 = (x^2)^1/2⋅(x^1−1)

2. Apply the power rule to the first term: (x^2)^1/2⋅x−1 = (1/2)x^(2+1)−x^1

3. Apply the power rule to the second term: (1/2)x^(2+1)−x^1 = (1/2)x^3−x^1

4. Integrate the function: (1/2)x^3−x^1 = (1/2)x^4−x^2 + C

Question:

Integrate the function xtan^−1x

Answer:

1. Rewrite the function as tan^-1(x) / x

2. Use the formula for integration of the form (1/u) du = ln|u| + c

3. Integrate tan^-1(x) / x = ln|x tan^-1(x)| + c

Question:

Integrate the function 1/√(x−1)(x−2)

Answer:

Answer:

1. First, rewrite the function in terms of the standard form of a fraction: 1/√(x−1)(x−2) = (1/√(x−1))/(x−2)

2. Use the power rule to expand the denominator: (1/√(x−1))/(x−2) = (1/√(x−1))/(x^2-2x)

3. Use the power rule to expand the numerator: (1/√(x−1))/(x^2-2x) = (1/x√(x−1))/(x^2-2x)

4. Multiply the numerator and denominator by √(x−1) to clear the fraction: (1/x√(x−1))/(x^2-2x) = (1/x^2(x−1))/(x^2-2x)√(x−1)

5. Integrate both sides using the substitution u = x−1: ∫(1/x^2(x−1))/(x^2-2x)√(x−1)dx = ∫(1/u^2)du

6. Solve the integral on the right side: ∫(1/u^2)du = -1/u

7. Substitute u = x−1 back into the equation: ∫(1/x^2(x−1))/(x^2-2x)√(x−1)dx = -1/(x−1)

8. The final answer is: ∫(1/√(x−1)(x−2))dx = -1/(x−1)

Question:

​Integrate the function 1/√9x^2+6x+5

Answer:

Answer:

Step 1: Rewrite the function as: 1/√(9x^2 + 6x + 5)

Step 2: Multiply the denominator by the conjugate of the denominator to get a perfect square: 1/[√(9x^2 + 6x + 5) * √(9x^2 + 6x + 5)]

Step 3: Rewrite the function as: 1/[(9x^2 + 6x + 5)^2]

Step 4: Rewrite the function as: 1/[9(x^2 + 2/3x + 5/9)]

Step 5: Rewrite the function as: 1/(9x^2 + 2/3x + 5/9)

Step 6: Integrate using the power rule: -1/3 (x^3 + 2/3x^2 + 5/9x) + C

Question:

​Find an anti derivative (or integral) of the given function by the method of inspection.cos3x

Answer:

Answer:

Step 1: Rewrite the given function as a sum of sines and cosines. cos3x = 4cos^3x - 3cosx

Step 2: Use the formula for anti-derivatives of sums of sines and cosines. ∫cos3x dx = ∫(4cos^3x - 3cosx) dx

Step 3: Integrate the individual terms. ∫cos3x dx = 4∫cos^3x dx - 3∫cosx dx

Step 4: Use the formula for the anti-derivative of a cosine cubed. ∫cos3x dx = 4(1/4sin^4x) - 3(sin x) + C

Step 5: Simplify the expression. ∫cos3x dx = sin^4x - 3sin x + C

Question:

Find an anti derivative (or integral) of the given function by the method of inspection.sin2x

Answer:

1. Rewrite the given function as 2sin xcos x.

2. Use the identity sin2A = 2sin A cos A to rewrite the function as sin2x.

3. Use the identity sin2A = 1 - cos2A to rewrite the function as 1 - cos2x.

4. Use the identity cos2A = (1 + cos2A)/2 to rewrite the function as (1 + cos2x)/2.

5. Integrate both sides with respect to x.

6. On the left side, the integral of 1 is x. On the right side, the integral of (1 + cos2x)/2 is (x + sin2x)/2.

7. The anti derivative of sin2x is (x + sin2x)/2.

Question:

Integrate the rational function 1/(e^x−1)

Answer:

Answer:

1. Rewrite the function as 1/e^x - 1/1
2. Integrate 1/e^x using the formula ∫1/e^x dx = -ln(e^x)
3. Integrate 1/1 using the formula ∫1/1 dx = x
4. Add the two integrals together to get the answer -ln(e^x) + x

Question:

Integrate the rational function x/(x−1)(x−2)(x−3)

Answer:

1. Rewrite the rational function as a sum of partial fractions: x/(x−1)(x−2)(x−3) = A/(x−1) + B/(x−2) + C/(x−3)

2. To solve for A, B, and C, use the following equation: A(x−2)(x−3) + B(x−1)(x−3) + C(x−1)(x−2) = x

3. Substitute x = 1, 2, and 3 into the equation to solve for A, B, and C: A(2)(3) + B(1)(3) + C(1)(2) = 1 6A + 3B + 2C = 1

4. Solve for A, B, and C: A = 1/6 B = -1/2 C = 1/3

5. Substitute A, B, and C into the original equation: x/(x−1)(x−2)(x−3) = 1/6(1/(x−1)) - 1/2(1/(x−2)) + 1/3(1/(x−3))

6. Simplify: x/(x−1)(x−2)(x−3) = 1/(6(x−1)) - 1/(2(x−2)) + 1/(3(x−3))

7. Integrate both sides of the equation: ∫x/(x−1)(x−2)(x−3) dx = ∫1/(6(x−1)) - 1/(2(x−2)) + 1/(3(x−3)) dx

8. Solve the integrals: ∫x/(x−1)(x−2)(x−3) dx = 1/6ln|x−1| - 1/2ln|x−2| + 1/3ln|x−3| + C

Question:

Integrate the rational function 1/x(x^n+1)

Answer:

1. Use the substitution u = x^n + 1.

2. Integrate 1/u du.

3. Substitute du back in terms of dx.

4. Integrate 1/x dx.

5. Substitute x^n + 1 back in terms of u.

6. Simplify the expression.

Question:

Integrate the function 2x/1+x^2

Answer:

1. First, use the substitution u = x^2, so that du = 2x dx

2. Next, rewrite the integral as ∫2x/1+x^2 dx = ∫2u/(1+u) du

3. Finally, integrate both sides of the equation to get: ∫2u/(1+u) du = (2/3)ln|1+u| + C

Question:

Find the integral of ∫x^2(1−1​/x^2)dx

Answer:

1. ∫x^2(1−1​/x^2)dx
2. ∫x^2 - 1/x^2 dx
3. ∫x^2dx - ∫1/x^2dx
4. (1/3)x^3 - (1/x) + C

Question:

Integrate the function 5x−2/1+2x+3x^2

Answer:

Answer: Step 1: Rewrite the function as 5x/(1 + 2x + 3x^2)

Step 2: Use partial fractions to separate the function into two fractions: 5x/(1 + 2x + 3x^2) = (A/1 + 2x) + (B/3x^2)

Step 3: Solve for A and B by multiplying both sides of the equation by (1 + 2x + 3x^2): 5x = A(1 + 2x + 3x^2) + B(1 + 2x)

Step 4: Solve for A and B by equating the coefficients of each term: A = 5 B = -2

Step 5: Substitute A and B into the original equation: 5x/(1 + 2x + 3x^2) = (5/1 + 2x) + (-2/3x^2)

Step 6: Integrate both sides of the equation: ∫5x/(1 + 2x + 3x^2) dx = ∫(5/1 + 2x) + (-2/3x^2) dx

Step 7: Integrate each side of the equation separately: ∫5x/(1 + 2x + 3x^2) dx = (5/2)ln|1 + 2x| + (2/3)ln|3x^2| + C, where C is the constant of integration.

Question:

Integrate the function 2cosx−3sinx/6cosx+4sinx

Answer:

1. Rewrite the function as (2cosx−3sinx)/(6cosx+4sinx)

2. Multiply the numerator and denominator by the conjugate of the denominator: (2cosx−3sinx)(6cosx−4sinx)/(6cosx+4sinx)(6cosx−4sinx)

3. Expand and simplify the numerator: 12cos^2x−20cosxsinx−18sinx^2

4. Rewrite the denominator as 36cos^2x−16sinx^2

5. Divide the numerator by the denominator: (12cos^2x−20cosxsinx−18sinx^2)/(36cos^2x−16sinx^2)

6. Use the trigonometric identity sin^2x+cos^2x=1 to simplify the denominator: (12cos^2x−20cosxsinx−18sinx^2)/(36−16sin^2x)

7. Use the trigonometric identity 1−sin^2x=cos^2x to simplify the denominator: (12cos^2x−20cosxsinx−18sinx^2)/(36cos^2x)

8. Integrate the function: (1/36)∫(12cos^2x−20cosxsinx−18sinx^2)dx

9. Integrate each term separately: (1/36)∫12cos^2xdx−(1/36)∫20cosxsinxdx−(1/36)∫18sinx^2dx

10. Solve each integral: (1/36)(3sin x−10cos x+6sinxcosx)+C

Question:

Find the integrals of the functions sin^4x

Answer:

Answer:

1. Rewrite the function as (sin^2x)^2
2. Use the formula for the integral of (u^n) = (u^(n+1))/(n+1), where n is a constant
3. Integrate (sin^2x)^2 = (sin^4x)/4 + C, where C is a constant

Question:

Integrate the function x+2/√x^2−1

Answer:

Step 1: Separate the function into two parts: x and 2/√x^2−1

Step 2: Integrate the first part, x, which is x^2/2

Step 3: Integrate the second part, 2/√x^2−1, which is 2*arcsin(x)

Question:

Find the integral of ∫√x(3x^2+2x+3)dx

Answer:

Answer:

1. Rewrite the integrand as a single fraction: ∫(3x^3+2x^2+3x)√x dx

2. Use the substitution u = x^2: ∫(3u^2+2u+3)√u du

3. Rewrite the integrand as a single fraction: ∫(3u^2+2u+3)du

4. Integrate: (3/5u^5/2)+ (1/2u^2)+ (3u) + C

Question:

∫dx​/sin^2xcos^2x equals A tanx+cotx+C B tanx−cotx+C C tanxcotx+C D tanx−cot2x+C

Answer:

Answer: A

Step-by-step solution:

1. Begin by using the identity sin2x + cos2x = 1
2. ∫dx/sin2xcos2x = ∫dx/(1 - sin2x)
3. Use the substitution u = sinx
4. du = cosx dx
5. ∫dx/(1 - sin2x) = ∫du/(1 - u2)
6. Integrate: ∫du/(1 - u2) = tan-1u + C
7. Substitute back: tan-1(sinx) + C = tanx + C

Question:

Integrate the rational function cosx/(1−sinx)(2−sinx)

Answer:

1. Rewrite the function as a sum of two fractions: cosx/(1−sinx) + cosx/(2−sinx)

2. Use the substitution u = sinx: cosx/(1−u) + cosx/(2−u)

3. Integrate each fraction separately: ∫cosx/(1−u) du = ln|1−u| + C1 ∫cosx/(2−u) du = ln|2−u| + C2

4. Substitute u = sinx back into the integrals: ln|1−sinx| + C1 + ln|2−sinx| + C2

Question:

Integrate the function 6x+7/√(x−5)(x−4)

Answer:

1. Rewrite the function as 6x+7/[(x-5)(√x-4)]
2. Use the product rule to rewrite as 6/√(x-4) + 7(√x-4)/[(x-5)(√x-4)]
3. Integrate the first term: ∫6/√(x-4)dx = 3√(x-4) + C
4. Integrate the second term: ∫7(√x-4)/(x-5)dx = 7ln|x-5| + 7√x-4 + C
5. Combine the two results: 3√(x-4) + 7ln|x-5| + 7√x-4 + C

Question:

Integrate the rational function 1−x^2/x(1−2x)

Answer:

Answer:

1. Rewrite the rational function as: (1 - x^2)/x - (2x - x^2)/x

2. Simplify: 1/x - 2 + x

3. Integrate: ln|x| - 2x + (1/2)x^2 + C

Question:

Integrate the rational function 2/(1−x)(1+x^2)

Answer:

Step 1: Rewrite the rational function as the sum of two fractions: 2/(1-x) + 2/(1+x^2)

Step 2: Integrate each fraction separately: ∫2/(1-x)dx = 2 ln|1-x| + C

Step 3: Integrate the second fraction: ∫2/(1+x^2)dx = 2 tan^-1(x) + C

Step 4: Combine the two integrals: 2 ln|1-x| + 2 tan^-1(x) + C

Question:

Integrate the rational function (x^2+1)(x^2+2)/(x^2+3)(x^2+4)

Answer:

Answer: Step 1: Rewrite the function in partial fraction form: (x^2+1)(x^2+2) / (x^2+3)(x^2+4) = A/(x^2+3) + B/(x^2+4) + C/(x^2+1) + D/(x^2+2)

Step 2: Solve for the constants A, B, C, and D: A(x^2+4) + B(x^2+3) + C(x^2+2) + D(x^2+1) = (x^2+1)(x^2+2)

A = 1, B = -1, C = 1, D = -1

Step 3: Substitute the constants back into the partial fraction form: 1/(x^2+3) - 1/(x^2+4) + 1/(x^2+1) - 1/(x^2+2)

Step 4: Integrate: ln|x^2+3| - ln|x^2+4| + ln|x^2+1| - ln|x^2+2| + C

Question:

Integrate the function : x√x+2

Answer:

1. Rewrite the function as x^(3/2) + 2

2. Integrate x^(3/2) using the power rule: (2/5)x^(5/2)

3. Add the constant of integration: (2/5)x^(5/2) + C

Question:

Integrate the rational function 3x−1/(x+2)^2

Answer:

Answer: Step 1: Use the formula for integration of rational functions: ∫(P(x)/Q(x))dx = (P(x)∫(1/Q(x))dx) + C

Step 2: Integrate the denominator: ∫(1/(x+2)^2)dx = -(1/(x+2)) + C

Step 3: Substitute the denominator into the original equation: ∫(3x−1/(x+2)^2)dx = (3x−1)(-(1/(x+2))) + C

Step 4: Simplify the equation: ∫(3x−1/(x+2)^2)dx = -(3x−1)/(x+2) + C

Question:

Integrate the rational function 1/x(x^4−1)

Answer:

1. ∫1/x(x^4−1)dx

2. ∫1/x dx + ∫x^4−1 dx

3. ln|x| + (1/5)x^5 - x

4. ln|x| + (1/5)(x^5 - x)

Question:

Integrate the function (logx)^2/x

Answer:

Step 1: Use the formula ∫ (logx)^2/x dx = ∫ (logx)^2 * d(1/x)

Step 2: Rewrite the equation as ∫ (logx)^2 * (-1/x^2) dx

Step 3: Use the substitution u = logx, du = 1/x dx

Step 4: Rewrite the equation as ∫ (-u^2) * (-1/x^2) du

Step 5: Integrate using the power rule ∫ (-u^2) * (-1/x^2) du = (-1/x^2) * (-1/3)u^3 + C

Step 6: Substitute back for u and x to get the final answer: (-1/3)(logx)^3 + C

Question:

Integrate the function xsin^−1x

Answer:

1. Rewrite the function as sin^-1(x) / x

2. Use the integration formula for inverse trigonometric functions: ∫sin^-1(x) / x dx = xsin^-1x - ∫dx/cos(sin^-1(x))

3. Use the integration formula for 1/cos(u): ∫dx/cos(sin^-1(x)) = ln|cos(sin^-1(x))| + C

4. Substitute the result from step 3 into step 2: xsin^-1x - ln|cos(sin^-1(x))| - C

5. The final result is: ∫xsin^−1x dx = xsin^-1x - ln|cos(sin^-1(x))| + C

Question:

Integrate the function (x−3)e^x/(x−1)^3

Answer:

1. Rewrite the function using partial fractions:

(x−3)e^x/(x−1)^3 = [A/(x-1)] + [B/(x-1)^2] + [C/(x-1)^3] + (x-3)e^x

2. Solve for A, B and C using the following equations:

A + B + C = 0 B + 2C = 3 C = -3

3. Substitute A, B and C into the equation and simplify:

(x−3)e^x/(x−1)^3 = [-1/(x-1)] + [3/(x-1)^2] + [-3/(x-1)^3] + (x-3)e^x

4. Integrate both sides of the equation:

∫ (x−3)e^x/(x−1)^3 dx = ∫ [-1/(x-1)] + [3/(x-1)^2] + [-3/(x-1)^3] + (x-3)e^x dx

5. Solve the integrals on the right side:

∫ (x−3)e^x/(x−1)^3 dx = ln|x-1| - (x-1) - (3/2)(x-1)^2 - (1/4)(x-1)^3 + (1/2)(x-3)e^x + C

6. Simplify the result:

∫ (x−3)e^x/(x−1)^3 dx = ln|x-1| - (3/2)(x-1)^2 - (1/4)(x-1)^3 + (1/2)(x-3)e^x + C

Question:

Integrate the function e^x(sinx+cosx)

Answer:

1. Integrate e^x = e^x + C

2. Integrate sinx + cosx = sin2x/2 + C

Question:

Integrate the function x^2/√x^6+a^6

Answer:

Answer:

1. Rewrite the function as x^2/(x^3√(x^3+a^6))
2. Take the integral of both sides: ∫x^2/(x^3√(x^3+a^6))dx
3. Use the substitution u = x^3 + a^6 ∫x^2/√u du
4. Integrate both sides: (1/3)x^3√u - (1/3)a^6√u + C
5. Substitute back in u = x^3 + a^6 (1/3)x^3√(x^3+a^6) - (1/3)a^6√(x^3+a^6) + C

Question:

Find the integrals of the functions tan^4x

Answer:

1. Integrate tan^4x using the power rule: ∫tan^4x dx = ∫x^4tan dx

2. Use the chain rule to rewrite the integral: ∫x^4tan dx = x^4 ∫tan dx

3. Integrate tan using the substitution u = tanx: ∫x^4tan dx = x^4 ∫u du

4. Integrate u using the power rule: ∫x^4tan dx = x^4 ∫u du = x^4(u^2/2 + C)

5. Substitute u = tanx back into the equation: ∫x^4tan dx = x^4(tan^2x/2 + C)

6. Simplify the equation: ∫tan^4x dx = (tan^6x/6) + C

Question:

Integrate the function √tanx/sinxcosx

Answer:

1. Integrate 1/cosx by using the substitution u = cosx: ∫1/cosx dx = ∫du/u = ln|u| + C = ln|cosx| + C

2. Integrate √tanx by using the substitution u = tanx: ∫√tanx dx = ∫√u du = (2/3)u^(3/2) + C = (2/3)tanx^(3/2) + C

3. Multiply the two integrals together: ∫√tanx/cosx dx = (2/3)tanx^(3/2)/cosx + C

Question:

Integrate the function sinx/1+cosx

Answer:

1. Rewrite the function as sinx/(1+cosx) = (sinx*(1-cosx))/(1+cosx)

2. Use the formula (u*v’)/(v) = u - (u/v)*v'

3. Substitute u = sinx and v = 1+cosx, then v’ = -sinx

4. Therefore, (sinx*(1-cosx))/(1+cosx) = sinx - (sinx/(1+cosx))(-sinx)

5. Integrate both sides with respect to x:

∫(sinx*(1-cosx))/(1+cosx)dx = ∫sinxdx - ∫(sinx/(1+cosx))(-sinx)dx

6. On the left side, use the formula ∫(u/v)dv = u*ln|v| + c

7. Therefore, sinx*ln|1+cosx| + c = ∫sinxdx - ∫(sinx/(1+cosx))(-sinx)dx

8. On the right side, use the formula ∫udv = uv + c

9. Therefore, sinxln|1+cosx| + c = xsinx + c - (sinxln|1+cosx| + c)

10. Simplify to get the answer: xsinx + c - sinxln|1+cosx| + c = xsinx - sinxln|1+cosx| + c

Question:

Find the integrals of the function tan^32xsec2x

Answer:

Answer: Step 1: Rewrite the function as (sec x tan^3 x)^2. Step 2: Use the chain rule to differentiate the function. Step 3: Integrate the function using the power rule. Step 4: The integral of (sec x tan^3 x)^2 is (1/4) sec x tan^4 x + C.

Question:

∫dx​/x(x^2+1) equals
A log∣x∣−1​/2log(x^2+1)+C B log∣x∣+1​/2log(x^2+1)+C C −log∣x∣+1​/2log(x^2+1)+C D 1​/2log∣x∣+log(x^2+1)+C

Answer:

Answer: D 1/2log∣x∣+log(x^2+1)+C

Question:

Integrate the function sin^−1(2x​/1+x^2)

Answer:

1. Let u = sin^−1(2x​/1+x^2)

2. du = (2/1+x^2)dx

3. ∫du = ∫(2/1+x^2)dx

4. ∫du = 2∫1/(1+x^2)dx

5. ∫du = 2∫(1/x^2+1)dx

6. ∫du = 2∫1/x^2dx + 2∫1dx

7. ∫du = 2∫1/x^2dx + 2x + C

8. ∫du = 2ln|x| + 2x + C

Question:

Integrate the function √x^2+4x−5

Answer:

1. First, rewrite the function in the form of a polynomial: x^2 + 4x - 5 = (x + 5)(x - 1).

2. Now, use the power rule for integration to calculate the integral: ∫(x + 5)(x - 1)dx = ∫x^2 - 1dx.

3. Use the power rule again to calculate the integral: ∫x^2 - 1dx = 1/3x^3 - x + C.

4. Finally, substitute the original function back in: ∫√x^2 + 4x - 5dx = 1/3x^3 - x + C.

Question:

∫ √x^2−8x+7 dx is equal to A 1​/2(x−4)√x^2−8x+7+9log∣x−4+√x^2−8x+7∣+C B 1​/2(x+4)√x^2−8x+7+9log∣x+4+√x^2−8x+7∣+C C 1​/2(x−4)√x^2−8x+7−32​log∣x−4+√x^2−8x+7∣+C D 1​/2(x−4)√x^2−8x+7−29​log∣x−4+√x^2−8x+7​∣+C

Answer:

Answer: A 1/2(x−4)√x^2−8x+7+9log∣x−4+√x^2−8x+7∣+C

Question:

Find the integrals of the functions sin^2(2x+5)

Answer:

Answer:

1. Integrate both sides with respect to x: ∫sin2(2x + 5)dx

2. Use the double angle formula to rewrite the integrand: ∫(1 - cos(4x + 10))/2 dx

3. Split the integrand into two integrals: ∫1/2 dx + ∫(cos(4x + 10))/2 dx

4. Integrate the first integral: 1/2x + C

5. Integrate the second integral using the trigonometric substitution u = 4x + 10: 1/2 ∫cosudu

6. Integrate the second integral using the trigonometric substitution u = 4x + 10: 1/2 sin(4x + 10) + C

Question:

Find the integral of ∫x^3+5x^2−4​/x^2dx

Answer:

Answer: Step 1: Use the method of partial fractions to rewrite the integrand as

∫(x^3+5x^2−4)/x^2dx = ∫x + 5 − 4/x dx

Step 2: Integrate both sides of the equation

∫x + 5 − 4/x dx = ∫x dx + ∫5 dx − ∫4/x dx

Step 3: Evaluate the integrals on the right side

∫x + 5 − 4/x dx = (1/2)x^2 + 5x − 4ln(x) + C

Question:

Find the integral of ∫2−3sinx​/cos^2xdx

Answer:

Answer: Step 1: Rewrite the integral as ∫2−3sinxdcosxdx.

Step 2: Use integration by parts, with u = sinx and dv = cosx dx.

Step 3: du = cosx dx and v = -sinx.

Step 4: Integrate u and v to get ∫2−3sinxdcosxdx = -sinxcosx|2−3 + ∫2−3cos^2xdx.

Step 5: Use the identity cos^2x = 1 - sin^2x to rewrite the integral.

Step 6: ∫2−3cos^2xdx = ∫2−3(1-sin^2x)dx.

Step 7: Use integration by parts again, with u = 1 and dv = -sin^2x dx.

Step 8: du = 0 dx and v = -1/2cos2x.

Step 9: Integrate u and v to get ∫2−3cos^2xdx = -1/2cos2x|2−3 + ∫2−3sin^2xdx.

Step 10: Use the identity sin^2x = 1 - cos^2x to rewrite the integral.

Step 11: ∫2−3sin^2xdx = ∫2−3(1-cos^2x)dx.

Step 12: Use integration by parts one last time, with u = 1 and dv = -cos^2x dx.

Step 13: du = 0 dx and v = 1/2sin2x.

Step 14: Integrate u and v to get ∫2−3sin^2xdx = 1/2sin2x|2−3 + ∫2−3cos^2xdx.

Step 15: Substitute the result from Step 14 into Step 9 to get ∫2−3cos^2xdx = -1/2cos2x|2−3 + 1/2sin2x|2−3 + ∫2−3cos^2xdx.

Step 16: Solve for ∫2−3cos^2xdx to get ∫2−3cos^2xdx = (1/2sin2x - 1/2cos2x)|2−3.

Step 17: Substitute the result from Step 16 into Step 4 to get ∫2−3sinxdcosxdx = -sinxcosx|2−3 + (1/2sin2x - 1/2cos2x)|2−3.

Step 18: Simplify the result to get the final answer: ∫2−3sinx​/cos^2xdx = -sinxcosx + 1/2sin2x - 1/2cos2x|2−3.

Question:

Integrate the function : √ax+b

Answer:

1. First, use the substitution rule to rewrite the equation as u = √ax + b.

2. Next, take the derivative of both sides of the equation to obtain du/dx = (1/2)a^(1/2)x^(-1/2).

3. Then, use the substitution rule again to rewrite the equation as du = (1/2)a^(1/2)x^(-1/2)dx.

4. Finally, integrate both sides of the equation to obtain ∫du = (1/2)a^(1/2)∫x^(-1/2)dx.

5. Solving the integral on the right side yields 2a^(1/2)x^(1/2) + c, where c is an arbitrary constant.

6. Substituting u back in yields 2a^(1/2)(√ax + b) + c.

Question:

The anti derivative of (√x+1​/√x) equals A 1​/3x^1/3​+2x^1​/2+C B 2​/3x^2/3​+1/2​x^2+C C 2​/3x^3/2​+2x^1​/2+C D 3​/2x^3/2​+1/2​x^1​/v+C

Answer:

Answer: C

Question:

Integrate the function 4x+1/√2x^2+x−3

Answer:

1. Rewrite the function as: 4x + (1/√2x^2) + x - 3

2. Add and subtract the terms: 4x + (1/√2x^2) + x - 3 = 4x + (1/√2x^2) + x + 3 - 3

3. Use the power rule to integrate: ∫4x + (1/√2x^2) + x + 3 - 3 dx

4. Apply the power rule: 2x^2 + x^2 + x^2 + 3x - 3x

5. Simplify: 4x^2 + 4x - 3x

6. Rewrite the function: ∫4x^2 + 4x - 3x dx

7. Apply the power rule: (2/3)x^3 + (2/2)x^2 - (3/2)x + C

Question:

Integrate the function x/9−4x^2

Answer:

1. ∫ x/9 − 4x² dx

2. ∫ x/9 dx − ∫ 4x² dx

3. (1/9)x²/2 − (2/3)x³

4. (1/18)x² − (2/9)x³ + C

Question:

Find the integral of ∫(2x^2+e^x)dx

Answer:

Step 1: Rewrite the integral in standard form by moving the constant term to the right side of the equation:

∫2x^2dx + ∫e^xdx

Step 2: Use the power rule to integrate the first term:

2/3x^3 + ∫e^xdx

Step 3: Use the integration rule for exponential functions to integrate the second term:

2/3x^3 + e^x

Step 4: Combine the two terms to get the final answer:

2/3x^3 + e^x

Question:

Integrate the function (4x+2)√x^2+x+1

Answer:

1. Rewrite the integrand as a perfect square: (4x+2)√(x+1)^2

2. Rewrite the integrand as a single term: (4x+2)(x+1)

3. Use the power rule to integrate: 1/2(4x+2)(x+1)^2 + C

Question:

Find the integral of ∫x^3−x^2+x−1​/x−1dx

Answer:

1. Rewrite the integral as ∫(x^3−x^2+x−1)dx/ (x−1)

2. Add 1 to both sides of the equation to get ∫(x^3−x^2+x)dx/ (x−1) + ∫1dx/(x−1)

3. Integrate the first term on the left-hand side to get (1/4)x^4 − (1/3)x^3 + (1/2)x^2 + x + C

4. Integrate the second term on the left-hand side to get ln|x−1| + C

5. Combine the two constants of integration to get (1/4)x^4 − (1/3)x^3 + (1/2)x^2 + x + ln|x−1| + C

Question:

∫ cos2xcos4xcos6xdx

Answer:

1. Use the formula ∫Acosmxcosnxdx = (A/2)(sinmxsinⁿ⁺¹x)/(mⁿ⁺¹ - nⁿ⁺¹)

2. ∫ cos2xcos4xcos6xdx = (1/2)(sin2xsin5x - sin4xsin7x)/(3 - 5)

3. (1/2)(sin2xsin5x - sin4xsin7x)/(3 - 5) = (1/2)(2sinxcos^4x - 4cos^2xsinxcos^3x)/(-2)

4. (1/2)(2sinxcos^4x - 4cos^2xsinxcos^3x)/(-2) = -(sinxcos^4x - 2cos^2xsinxcos^3x)

Question:

Find the integrals of the functions 1−cosx/1+cosx

Answer:

Answer: Step 1: Rewrite the function as (1-cosx)/(1+cosx)

Step 2: Multiply the numerator and denominator by the conjugate of the denominator, which is (1-cosx)

Step 3: Simplify the function to (1-cosx)^2/(1+cosx)^2

Step 4: Use the substitution u = 1+cosx

Step 5: Rewrite the function as (1-cosx)^2/u^2

Step 6: Integrate the function with respect to x

Step 7: The integral of (1-cosx)^2/u^2 with respect to x is (1-cosx)^2*ln(u)+C

Step 8: Substitute u back with 1+cosx

Step 9: The integral of (1-cosx)/(1+cosx) with respect to x is (1-cosx)^2*ln(1+cosx)+C

Question:

Integrate the rational function 3x+5/x^3−x^2−x+1

Answer:

1. Rewrite the rational function as: 3x + 5/(x - 1)(x^2 + x + 1)

2. Use the formula for integration of a rational function by partial fractions: ∫(3x + 5/(x - 1)(x^2 + x + 1))dx

3. Separate the rational function into partial fractions: ∫(A/(x - 1) + Bx + C/(x^2 + x + 1))dx

4. Solve for A, B, and C: A = 3, B = 5, C = -3

5. Substitute the values for A, B, and C into the equation: ∫(3/(x - 1) + 5x - 3/(x^2 + x + 1))dx

6. Integrate the partial fractions: 3ln|x - 1| + 5x - 3/2(2x + 1)ln|x^2 + x + 1| + C

Question:

Integrate the function x(logx)^2

Answer:

Step 1: Use the substitution rule to rewrite x(logx)^2 as u^2, where u = logx.

Step 2: Integrate u^2 with respect to x.

Step 3: ∫u^2dx = (1/3)u^3 + c

Step 4: Substitute u = logx back into the equation.

Step 5: (1/3)logx^3 + c

Question:

Integrate the function e^2x−e^−2x/e^2x+e^−2x

Answer:

1. Rewrite the function as: (e^2x - e^-2x) / (e^2x + e^-2x)

2. Multiply both numerator and denominator by e^2x: (e^4x - 1) / (e^4x + 1)

3. Rewrite the numerator as: (e^4x - 1) = (e^2x + 1)(e^2x - 1)

4. Rewrite the denominator as: (e^4x + 1) = (e^2x + 1)(e^2x + 1)

5. Rewrite the function as: (e^2x + 1)(e^2x - 1) / (e^2x + 1)(e^2x + 1)

6. Factor out the common term in the numerator and denominator: (e^2x + 1)(e^2x - 1) / (e^2x + 1)^2

7. Rewrite the function as: (e^2x + 1)(e^2x - 1) / (e^2x + 1)^2

8. Integrate: ∫ (e^2x + 1)(e^2x - 1) / (e^2x + 1)^2 dx

9. Use the formula ∫ udv = uv - ∫ vdu:

∫ (e^2x + 1)(e^2x - 1) / (e^2x + 1)^2 dx = (1/2)(e^2x - 1/e^2x) - (1/2)(∫ (2e^2x + 2) / (e^2x + 1)^2 dx)

10. Use the formula ∫ u/v dx = ∫ udx / v:

∫ (2e^2x + 2) / (e^2x + 1)^2 dx = (2/e^2x + 2) ∫ (1 / (e^2x + 1) dx)

11. Use the formula ∫ 1/u du = ln|u|:

∫ (1 / (e^2x + 1) dx) = ln|e^2x + 1| + C

12. Substitute back into the original equation:

∫ (e^2x + 1)(e^2x - 1) / (e^2x + 1)^2 dx = (1/2)(e^2x - 1/e^2x) - (1/2)(2/e^2x + 2) ln|e^2x + 1| + C

Question:

Find the integral of the function 1/sinxcos^3x

Answer:

Answer:

Step 1: Rewrite the function in terms of sin x and cos x:

1/sinxcos^3x = (cosx/sinx) * (1/cos^2x)

Step 2: Use the following integration formulas:

∫ (cosx/sinx) dx = ln|sinx| + c

∫ (1/cos^2x) dx = tanx + c

Step 3: Integrate the function:

∫ (cosx/sinx) * (1/cos^2x) dx = (ln|sinx| + c)*(tanx + c)

Step 4: Simplify the integrand:

∫ (cosx/sinx) * (1/cos^2x) dx = ln|sinx|*tanx + c^2

Question:

Integrate the function x^2/(2+3x^3)^3

Answer:

1. Let u = 2 + 3x^3
2. du = 9x^2dx
3. ∫x^2/(2+3x^3)^3dx = ∫x^2/u^3du
4. ∫x^2/u^3du = 1/2u^(-2) + C
5. ∫x^2/(2+3x^3)^3dx = 1/2(2+3x^3)^(-2) + C

Question:

Integrate the function 1/x+xlogx

Answer:

Answer:

1. Rewrite the function as: (1/x) + (xlnx)
2. Integrate both terms separately: ∫(1/x)dx = ln|x| + C ∫(xlnx)dx = (xlnx) - (x) + C
3. Add the two terms together: ln|x| + (xlnx) - (x) + C

Question:

Integrate the rational function 1/x^4−1

Answer:

Answer: Step 1: Rewrite the rational function as the sum of two fractions: 1/x^4 - 1 = (1/x^4) + (-1)

Step 2: Use the power rule to integrate each fraction: Integral (1/x^4) dx = -1/x^3 + C Integral (-1) dx = -x + C

Step 3: Add the two integrals together: -1/x^3 - x + C = -1/x^3 - x + C

Question:

Integrate the function e^2x+3

Answer:

Answer: Step 1: Rewrite the function as u = e2x + 3

Step 2: Take the antiderivative of u, which is ∫u du = ∫(e2x + 3) du

Step 3: Apply the power rule and simplify, which is 1/2*e2x + 3x + C, where C is the constant of integration.

Question:

Integrate the function: x^2/1−x^6

Answer:

1. Start by rewriting the function in terms of fractions: (x^2)/(1 - x^6)

2. Use the power rule to rewrite the denominator: (x^2)/(1 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)

3. Use the power rule to rewrite the numerator: (2x - 6x^3 + 10x^4 - 10x^5 + 4x^6)/(1 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)

4. Use the long division method to divide the numerator by the denominator: x - 3x^2 + 5x^3 - 5x^4 + 2x^5 + C

5. Add the constant of integration: x - 3x^2 + 5x^3 - 5x^4 + 2x^5 + C

Question:

Integrate the function √4−x^2

Answer:

1. ∫√4−x^2 dx

2. Let u = 4−x^2

3. du = -2x dx

4. ∫du = -1/2 ∫2x dx

5. ∫du = -1/2 x^2 + C

6. ∫√4−x^2 dx = -1/2 (4−x^2)^2 + C

Question:

Integrate the following function.sinxsin(cosx)

Answer:

Answer: Step 1: Use the formula ∫sinxsin(cosx)dx = -1/2cos(2x) + C

Step 2: Integrate -1/2cos(2x)

Step 3: -1/4sin(2x) + C

Question:

Integrate the function e2^x−1/e^2x+1

Answer:

1. Take the natural log of both sides: ln(e2^x−1/e^2x+1)

2. Use the power rule to expand the left-hand side: x ln(e2) - ln(e^2x+1)

3. Use the natural log rule to expand the right-hand side: x ln(e2) - ln(e^2x+1)

4. Simplify the equation: ln(e2^x−1/e^2x+1) = 0

5. Take the exponential of both sides: e2^x−1/e^2x+1 = e^0

6. Simplify the equation: e2^x−1/e^2x+1 = 1

7. Solve for x: e2^x−1/e^2x+1 = 1 e2^x−e^2x = 0 2^x(e−1) = 0 2^x = 0 x = 0

Question:

Integrate the function cosx/√1+sinx

Answer:

1. Rewrite the function as cosx/(1+sinx)^1/2

2. Use the substitution u = sinx

3. du = cosx dx

4. ∫ cosx/(1+sinx)^1/2 du = ∫ (1/2)u^-1/2 du

5. Integrate the right side of the equation using the power rule:

6. (1/2)u^1/2 + C = (1/2)sinx^1/2 + C

Question:

Find the integrals of the function cosx−sinx/1+sin2x

Answer:

Answer:

1. Rewrite the function in terms of sine and cosine: cosx−sinx/1+sin2x = (cosx - sinx)/(1 + sin^2x)

2. Use the trigonometric identity sin^2x + cos^2x = 1 to simplify the denominator: (cosx - sinx)/(1 + sin^2x) = (cosx - sinx)/(1 + (1 - cos^2x))

3. Use the trigonometric identity sin2x = 2sinxcosx to simplify the denominator: (cosx - sinx)/(1 + (1 - cos^2x)) = (cosx - sinx)/(1 + 2sinxcosx)

4. Use the trigonometric identity sin(A + B) = sinAcosB + cosAsinB to simplify the numerator: (cosx - sinx)/(1 + 2sinxcosx) = (cosxcosx + sinxsinx)/(1 + 2sinxcosx)

5. Use the trigonometric identity cos2x = cos^2x - sin^2x to simplify the numerator: (cosxcosx + sinxsinx)/(1 + 2sinxcosx) = (cos^2x - sin^2x)/(1 + 2sinxcosx)

6. Use the substitution u = sinxcosx to simplify the denominator: (cos^2x - sin^2x)/(1 + 2sinxcosx) = (cos^2x - sin^2x)/(1 + 2u)

7. Integrate both sides: ∫(cos^2x - sin^2x)/(1 + 2u)du = ∫1/(1 + 2u)du

8. Solve the integral on the left side: ∫(cos^2x - sin^2x)/(1 + 2u)du = 1/2 ln|1 + 2u| + C

9. Solve the integral on the right side: ∫1/(1 + 2u)du = 1/2 ln|1 + 2u| + C

10. Therefore, the integral of the function cosx−sinx/1+sin2x is: ∫(cosx−sinx/1+sin2x)dx = 1/2 ln|1 + 2sinxcosx| + C

Question:

Find the integrals of the functions sin^3xcos^3x

Answer:

1. Rewrite the function as (sinxcosx)^3.

2. Use the formula ∫u^n du = (u^(n+1))/(n+1) to integrate.

3. Integrate (sinxcosx)^3 to get (sinxcosx)^4/4.

Question:

Find the integral of the function sin^−1(cosx)

Answer:

Answer: Step 1: Rewrite the function as arcsin(cosx) Step 2: Use the integration formula ∫arcsin(x)dx = xarcsin(x) - √(1-x^2) + C Step 3: Substitute cosx for x Step 4: Integrate: ∫arcsin(cosx)dx = cosxarcsin(cosx) - √(1-cos^2x) + C

Question:

Find the integrals of the functions sin^3(2x+1)

Answer:

1. Use the power rule to rewrite the function as sin^2(2x+1) * sin(2x+1)
2. Integrate sin^2(2x+1) using the power rule: (1/3)sin^3(2x+1) + (1/2)sin^2(2x+1)cos(2x+1) + C
3. Integrate sin(2x+1) using the power rule: -(1/2)cos^2(2x+1) + (1/2)sin(2x+1)cos(2x+1) + C
4. Combine the two integrals: (1/3)sin^3(2x+1) + (1/2)sin^2(2x+1)cos(2x+1) - (1/2)cos^2(2x+1) + (1/2)sin(2x+1)cos(2x+1) + C

Question:

Integrate the function cotxlogsinx

Answer:

1. Use the formula for integration by parts: ∫u dv = uv - ∫v du

2. Let u = ln(sin x) and dv = cotx dx

3. Then du = (cos x)/(sin x) dx and v = -ln|cotx|

4. Substituting these values in the formula for integration by parts, ∫u dv = uv - ∫v du gives ∫ln(sin x) cotx dx = -ln|cotx| ln(sin x) - ∫-ln|cotx| (cos x)/(sin x) dx

5. Simplifying, ∫ln(sin x) cotx dx = -ln|cotx| ln(sin x) + ∫ln|cotx| (sin x)/(cos x) dx

6. Using the formula for integration by parts once again, ∫u dv = uv - ∫v du with u = ln|cotx| and dv = (sin x)/(cos x) dx, we get ∫ln|cotx| (sin x)/(cos x) dx = ln|cotx| (sin x) - ∫(sin x)/(cos x) ln|cotx| dx

7. Simplifying, ∫ln(sin x) cotx dx = -ln|cotx| ln(sin x) + ln|cotx| (sin x) - ∫(sin x)/(cos x) ln|cotx| dx

8. Finally, ∫ln(sin x) cotx dx = ln|cotx| (sin x) - ∫(sin x)/(cos x) ln|cotx| dx

Question:

Integrate the function sinx/(1+cosx)^2

Answer:

Answer: Step 1: Rewrite the function as sinx/(1-sinx^2)

Step 2: Use the substitution u = sinx

Step 3: Integrate 1/(1-u^2)

Step 4: Integral of 1/(1-u^2) = (1/2) * ln|1-u^2| + C

Step 5: Substitute u = sinx

Step 6: Integral of sinx/(1-sinx^2) = (1/2) * ln|1-sinx^2| + C

Question:

Integrate the function 1/√1+4x^2

Answer:

Answer: Step 1: Apply integration by parts.

Let u = 1/√1 + 4x^2 and dv = dx

du = -2x/√1 + 4x^2 dx

v = x

Step 2: Integrate both sides.

∫u dv = ∫v du

∫1/√1 + 4x^2 dx = x√1 + 4x^2 - ∫-2x/√1 + 4x^2 dx

Step 3: Solve the integral on the right side.

∫1/√1 + 4x^2 dx = x√1 + 4x^2 - (1/2)∫1/√1 + 4x^2 d(2x)

Step 4: Solve the integral on the right side.

∫1/√1 + 4x^2 dx = x√1 + 4x^2 - (1/2)(1/2)(2x)√1 + 4x^2 + C

Step 5: Simplify the expression.

∫1/√1 + 4x^2 dx = (1/2)x√1 + 4x^2 + C

Question:

Integrate the rational function 2x/(x^2+1)(x^2+3)

Answer:

Step 1: Start by breaking up the rational function into two separate fractions: 2x/(x^2+1) and (x^2+3).

Step 2: Integrate each fraction separately.

For 2x/(x^2+1): Integrate 2x/(x^2+1) = 2/2 * (x/x^2+1) = x/(x^2+1) + C

For (x^2+3): Integrate (x^2+3) = 1/3 * (x^3+3x) + C

Step 3: Add the two integrals together to get the final answer: Integrate 2x/(x^2+1)(x^2+3) = x/(x^2+1) + 1/3*(x^3+3x) + C

Question:

Integrate the function xsinx

Answer:

Answer:

1. Rewrite the function as sin(x) * x
2. Use the integration formula ∫sin(x)dx = -cos(x) + C
3. Use the integration by parts formula ∫uvdx = u∫vdx - ∫v∫udx
4. Let u = x and dv = sin(x)dx
5. Therefore, du = dx and v = -cos(x)
6. Plugging these values into the integration by parts formula: ∫xsin(x)dx = x(-cos(x)) - ∫-cos(x)dx
7. Use the integration formula ∫-cos(x)dx = sin(x) + C
8. Plugging this value back into the original equation: ∫xsin(x)dx = x(-cos(x)) - (sin(x) + C)
9. Simplify: ∫xsin(x)dx = x(-cos(x)) - sin(x) - C

Question:

Integrate the function e^tan−1x/1+x^2

Answer:

Step 1: Use the substitution u = tan−1x

Step 2: du = 1/(1+x^2) dx

Step 3: ∫ e^tan−1x/(1+x^2) dx = ∫ e^u du

Step 4: Integrate both sides, ∫ e^u du = e^u + C

Step 5: Substitute u = tan−1x back in, ∫ e^tan−1x/(1+x^2) dx = e^tan−1x + C

Question:

Find the integral of ∫(√x−1​/√x)^2dx

Answer:

Step 1: Rewrite the integral using the power rule: ∫(x^(1/2) - 1)^2dx

Step 2: Rewrite the integrand using the product rule: 2x^(1/2) - 2x^(1/2) + 2

Step 3: Integrate the integrand using the power rule: x^(3/2) - 2x^(1/2) + 2x + C

Question:

Find the integral of ∫(1+x)√xdx

Answer:

1. Use the substitution method: Let u = 1 + x, then du = dx

2. Rewrite the integral: ∫u√(u-1)du

3. Integrate: (2/3)u^(3/2) - (2/5)u^(5/2) + C

Question:

Find the integral of the function cos2x+2sin^2x/cos2x

Answer:

Answer: Step 1: Rewrite the function as 2sin^2x/cos2x + cos2x Step 2: Rewrite the function as sin2x + tan2x Step 3: Integrate both sides with respect to x Step 4: Integrate sin2x with respect to x: (1/2)cos2x Step 5: Integrate tan2x with respect to x: -(1/2)ln|cos2x| + C Step 6: Combine the two integrals: (1/2)cos2x - (1/2)ln|cos2x| + C

Question:

Integrate the function: x√1+2x^2

Answer:

1. Rewrite the function as: (x^2 + 2x^2)^(1/2)

2. Use the power rule to integrate: (2x^3/3)^(1/2)

3. Simplify: (2/3)x^(3/2)

Question:

Integrate the function x+2/√x^2+2x+3

Answer:

Step 1: Rewrite the equation as ∫(x+2)/(x^2+2x+3)dx

Step 2: Rewrite the denominator as (x+3)(x+1)

Step 3: Use the substitution u=x+3

Step 4: Rewrite the equation as ∫(u-1)/u du

Step 5: Integrate both sides to get

Step 6: ∫(u-1)/u du = ln|u| + C

Question:

Find the integrals of the functions cos^42x

Answer:

1. Rewrite the function as (cos^2x)^21: (cos^2x)^21

2. Use the power rule of integration: (1/21) (cos^2x)^22 + C

3. Simplify: (1/21) (cos^44x) + C

Question:

Find the integrals of the function sin3xcos4x

Answer:

Answer: Step 1: Rewrite the given function in terms of sine and cosine: sin3xcos4x = (1/2) [sin(7x) - sin(x)]

Step 2: Integrate both sides: (1/2) [sin(7x) - sin(x)] dx = (1/2) [-cos(7x) + cos(x)] + C

Step 3: Simplify the result: ∫sin3xcos4x dx = (1/2) [-cos(7x) + cos(x)] + C

Question:

Find the integral of ∫sec^2x​/cosec^2xdx

Answer:

Step 1: Use the substitution u = cosecx

Step 2: Differentiate u with respect to x to find du/dx: du/dx = -cosecxcotx

Step 3: Substitute u and du/dx into the original equation: ∫sec^2x(-cosecxcotx)dx

Step 4: Rewrite the equation as: -∫u^2du

Step 5: Integrate both sides of the equation: -1/3u^3 + C

Step 6: Substitute u back into the equation: -1/3(cosecx)^3 + C

Question:

If d​/dxf(x)=4x^3−3​/x^4 such that f(2)=0. Then f(x) is A x^4+1x^3​−129/8 ​B x^3+1x^4​+129/8 ​C x^4+1x^3​+129/8 ​D x^3+1x^4​−129/8

Answer:

Answer: C x^4+1x^3​+129/8