Question:

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

Answer:

Answer: Step 1: Rewrite the function in the form of a fraction: 1/(x^2 + 1) * (x^2 + 4)

Step 2: Use the formula for integration of a product of two functions: Integral of (f(x) * g(x)) = (Integral of f(x)) * (Integral of g(x))

Step 3: Integrate each function separately: Integral of 1/(x^2 + 1) = arctan(x) + C Integral of (x^2 + 4) = 1/3 x^3 + 4x + C

Step 4: Substitute the integrals into the formula: Integral of (1/(x^2 + 1) * (x^2 + 4)) = (arctan(x) + C) * (1/3 x^3 + 4x + C)

Step 5: Simplify the expression: Integral of (1/(x^2 + 1) * (x^2 + 4)) = 1/3 x^3 arctan(x) + 4x arctan(x) + C(1/3 x^3 + 4x + C)

Question:

Integrate the function cosx/√4−sin^2x

Answer:

1. Rewrite the function as cosx/(2√(1−sin^2x))

2. Use the substitution u = sinx

3. du = cosx dx

4. Integrate 1/(2√(1−u^2)) du

5. Integrate 1/2 * (1/√(1−u^2)) du

6. Integrate (1/2)*(arcsin(u)) + C

7. Integrate (1/2)*(arcsin(sinx)) + C

Question:

Integrate the function √x^2+1[log(x^2+1)−2logx]/x4

Answer:

Answer:

Step 1: Use the substitution u = x2 + 1

Step 2: Integrate (1/2) log u du

Step 3: Integrate (1/2) u1/2 du

Step 4: Integrate (2/3) u3/2

Step 5: Substitute back u = x2 + 1

Step 6: Final answer = (2/3)(x2 + 1)3/2

Question:

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

Answer:

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

Step 2: Rewrite the denominator as a product of two linear factors: (x + 2)(x + 1)

Step 3: Use the method of partial fractions to rewrite the expression as a sum of two fractions: A/(x + 2) + B/(x + 1)

Step 4: Use the method of undetermined coefficients to solve for A and B: A = 1/3 B = 1/3

Step 5: Substitute the values of A and B into the expression: 1/3/(x + 2) + 1/3/(x + 1)

Step 6: Use the Sum Rule of Integration to integrate the expression: ln|x + 2| - ln|x + 1| + C

Question:

Integrate the function: cos^3xe^logsin x

Answer:

Step 1: Use the substitution u = sin x

Step 2: d/dx cos^3x e^logsin x = cos^3x e^logsin x (3cos^2x (d/dx sin x) + e^logsin x (d/dx logsin x))

Step 3: d/dx sin x = cos x

Step 4: d/dx logsin x = 1/(sin x) (d/dx sin x) = 1/(sin x) cos x

Step 5: Substitute back in to get d/dx cos^3x e^logsin x = cos^3x e^logsin x (3cos^2x cos x + e^logsin x (1/(sin x) cos x))

Step 6: Integrate both sides to get ∫cos^3xe^logsin x dx = ∫ (cos^3x e^logsin x (3cos^2x cos x + e^logsin x (1/(sin x) cos x)) dx

Step 7: Integrate both sides to get ∫cos^3xe^logsin x dx = ∫cos^4x e^logsin x dx + ∫e^logsin x dx

Step 8: Use the substitution u = sin x for both integrals

Step 9: ∫cos^3xe^logsin x dx = 1/5 sin^5 x e^logsin x + sin x e^logsin x + C

Question:

Integrate the function 1/cos(x+a)cos(x+b)

Answer:

1. Rewrite the function as 1/cos(x+a)*1/sin(x+b): 1/cos(x+a)*1/sin(x+b)

2. Use the identity sin(x+b) = sin(x)cos(b) + cos(x)sin(b): 1/cos(x+a)*1/(sin(x)cos(b) + cos(x)sin(b))

3. Multiply the numerator and denominator by sin(x)cos(b) + cos(x)sin(b): (sin(x)cos(b) + cos(x)sin(b))/[cos(x+a)*(sin(x)cos(b) + cos(x)sin(b))]

4. Simplify the denominator using the identity cos(x+a) = cos(x)cos(a) - sin(x)sin(a): (sin(x)cos(b) + cos(x)sin(b))/[(cos(x)cos(a) - sin(x)sin(a))*(sin(x)cos(b) + cos(x)sin(b))]

5. Simplify the denominator using the identity sin(x)cos(b) + cos(x)sin(b) = sin(x+b): (sin(x)cos(b) + cos(x)sin(b))/[(cos(x)cos(a) - sin(x)sin(a))*sin(x+b)]

6. Simplify the numerator using the identity sin(x)cos(b) + cos(x)sin(b) = sin(x+b): sin(x+b)/[(cos(x)cos(a) - sin(x)sin(a))*sin(x+b)]

7. Integrate both sides with respect to x: ∫sin(x+b)/[(cos(x)cos(a) - sin(x)sin(a))*sin(x+b)]dx

8. Use the identity cos(x)cos(a) - sin(x)sin(a) = cos(x+a): ∫sin(x+b)/[cos(x+a)*sin(x+b)]dx

9. Use the identity sin(x+b) = sin(x)cos(b) + cos(x)sin(b): ∫[sin(x)cos(b) + cos(x)sin(b)]/[cos(x+a)*(sin(x)cos(b) + cos(x)sin(b))]dx

10. Use the identity sin(x)cos(b) + cos(x)sin(b) = sin(x+b): ∫sin(x+b)/[cos(x+a)*sin(x+b)]dx

11. Use the identity cos(x+a) = cos(x)cos(a) - sin(x)sin(a): ∫sin(x+b)/[(cos(x)cos(a) - sin(x)sin(a))*sin(x+b)]dx

12. Integrate: -1/cos(a)*ln|cos(x+a)| + C

Question:

∫cos2x​/(sinx+cosx)^2dx is equal to A −1​/(sinx+cosx)+C B log∣sinx+cosx∣+C C log∣sinx−cosx∣+C D 1/(sinx+cosx)^2

Answer:

1. Start by using the chain rule to rewrite the integrand as:

-2sin2x/(sinx+cosx)^2

2. Multiply the numerator and denominator by (sinx-cosx):

-2sin2x(sinx-cosx)/(sinx+cosx)(sinx-cosx)^2

3. Simplify the denominator:

-2sin2x(sinx-cosx)/(sinx-cosx)^3

4. Use the trigonometric identity sin2x = 2sinxcosx to rewrite the numerator:

-4sinxcosx(sinx-cosx)/(sinx-cosx)^3

5. Simplify the numerator:

-4sinx/ (sinx-cosx)^2

6. Rewrite the integrand in terms of u = sinx+cosx:

-4sinx/(u^2 - 2u + 1)

7. Integrate:

A -1/(u) + C

Question:

Integrate the function x^3/√1−x^8

Answer:

Step 1: Rewrite the function so that it is in the form f(x)dx f(x) = x^3 dx = √1-x^8

Step 2: Integrate both sides ∫f(x)dx = ∫x^3dx

Step 3: Apply the power rule to the left side (1/4)x^4 + C =

Step 4: Apply the substitution u = 1-x^8 to the right side (1/4)x^4 + C = (1/4)(1-u)^(4/8) + C

Step 5: Solve for x (1/4)x^4 + C = (1/4)(1-x^8)^(4/8) + C 0 = (1/4)(1-x^8)^(4/8) - (1/4)x^4

Step 6: Solve the equation for x 0 = (1/4)(1-x^8)^(4/8) - (1/4)x^4 x^4 = (1-x^8)^(4/8) x = (1-(1-x^8)^(4/8))^(1/4)

Question:

Integrate the function sin^8x−cos^8x/1−2sin^2xcos^2x

Answer:

1. Begin by rewriting the function as: sin^8x − cos^8x / (1 − 2sin^2xcos^2x)

2. Use the identity sin^2x + cos^2x = 1 to rewrite the denominator as: sin^8x − cos^8x / (1 − 2sin^2x(1 - sin^2x))

3. Use the identity sin^2x - cos^2x = -2sinxcosp to rewrite the denominator as: sin^8x − cos^8x / (1 − 2sinxcosp(2sinxcosp))

4. Use the identity sin^2xcos^2x = (1/2)(sin2x)^2 to rewrite the denominator as: sin^8x − cos^8x / (1 − (sin2x)^2)

5. Use the identity (sin2x)^2 = 1 - cos2x to rewrite the denominator as: sin^8x − cos^8x / (1 − 1 + cos2x)

6. Rewrite the entire fraction as: (sin^8x − cos^8x) / (cos2x)

7. Integrate both sides: ∫(sin^8x − cos^8x) / (cos2x)dx

8. Use the identity sin2x = 2sinxcosx to rewrite the numerator as: ∫(2sin^7xcosx − 2cos^7xsinx) / (cos2x)dx

9. Use the identity cos2x = 1 - sin2x to rewrite the denominator as: ∫(2sin^7xcosx − 2cos^7xsinx) / (1 - sin2x)dx

10. Use the identity sin2x = 2sinxcosx to rewrite the denominator as: ∫(2sin^7xcosx − 2cos^7xsinx) / (1 - 2sinxcosx)dx

11. Integrate both sides: ∫(2sin^7xcosx − 2cos^7xsinx)dx / (1 - 2sinxcosx) + C

Question:

Integrate the function e^5 log x−e^4 log x/e^3 log x−e^2 log x

Answer:

1. Rewrite the function as e^5 log x(e^-3 log x + e^-2 log x)- e^4 log x(e^-3 log x + e^-2 log x)

2. Use the product rule of integration: ∫u dv = uv - ∫v du

3. Let u = e^5 log x and dv = (e^-3 log x + e^-2 log x)dx

4. Then du = e^5 log x dx and v = -(1/e^3 log x + 1/e^2 log x)

5. ∫e^5 log x(e^-3 log x + e^-2 log x)dx = e^5 log x(-1/e^3 log x - 1/e^2 log x) - ∫-(1/e^3 log x + 1/e^2 log x)e^5 log x dx

6. ∫e^5 log x(e^-3 log x + e^-2 log x)dx = e^5 log x(-1/e^3 log x - 1/e^2 log x) + (1/e^8 log x + 1/e^7 log x) + C

Question:

Integrate the function: e^3logx(x^4+1)^−1

Answer:

1. Take the log of both sides: log(e^3logx(x^4+1)^−1)
2. Use the property of logarithms: log(e^3logx(x^4+1)^−1) = 3log(e) + log(logx) + log((x^4+1)^−1)
3. Use the power rule for logarithms: 3log(e) + log(logx) + log((x^4+1)^−1) = 3 + logx + (−1)(x^4+1)^−2*(4x^3)
4. Use the product rule for logarithms: 3 + logx + (−1)(x^4+1)^−2*(4x^3) = 3 + logx + (−1)(x^4+1)^−2*log(4x^3)
5. Use the power rule for logarithms: 3 + logx + (−1)(x^4+1)^−2log(4x^3) = 3 + logx + (−1)(x^4+1)^−2(4*logx + 3log(x))
6. Use the product rule for logarithms: 3 + logx + (−1)(x^4+1)^−2*(4logx + 3log(x)) = 3 + logx + (−1)(x^4+1)^−2log(4x^3)
7. Use the power rule for logarithms: 3 + logx + (−1)(x^4+1)^−2log(4x^3) = 3 + logx + (−1)(x^4+1)^−2(4*logx + 3log(x))
8. Integrate both sides: ∫3 + logx + (−1)(x^4+1)^−2*(4*logx + 3log(x))dx
9. Solve the integral: ∫3 + logx + (−1)(x^4+1)^−2*(4logx + 3log(x))dx = 3x + xlogx + (−1)(x^4+1)^−1(2logx² + 3xlogx) + C

Question:

Integrate the function (sin^−1√x−cos^−1√x/sin^−1√x+cos^1−√x)x∈[0,1]

Answer:

Solution: Step 1: Rewrite the function as (sin^−1√x−cos^−1√x/sin^−1√x+cos^−1√x)x

Step 2: Apply the power rule to the numerator and denominator = (sin^−2√x−cos^−2√x/sin^−2√x+cos^−2√x)x

Step 3: Rewrite the numerator and denominator in terms of sine and cosine = (1−cos2√x/1+cos2√x)x

Step 4: Apply the substitution u = cos2√x = (1−u/1+u)xdu

Step 5: Integrate = xln|1+u|−xln|1−u|+C

Step 6: Substitute u back to cos2√x = xln|1+cos2√x|−xln|1−cos2√x|+C

Step 7: Simplify = xln|sec2√x|+C

Question:

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

Answer:

Answer: Step 1: Rewrite the function as e^x/(1+e^x) + e^x/(1+e^x) Step 2: Integrate each term separately Step 3: Integrate e^x/(1+e^x) using u-substitution with u = 1 + e^x Step 4: Integrate e^x/(1+e^x) = ln|u| + C Step 5: Integrate e^x/(1+e^x) using u-substitution with u = 2 + e^x Step 6: Integrate e^x/(1+e^x) = ln|u| + C Step 7: Add the two integrals together: ln|u| + ln|u| + 2C = ln|u^2| + 2C

Question:

∫ dx​/e^x+e^−x is equal to A tan^−1(e^x)+C B tan^−1(e^−x)+C C log(e^x−e^−x)+C D log(e^x+e^−x)+C

Answer:

Answer: D log(e^x+e^−x)+C

Question:

Integrate the function sinx/sin(x−a)

Answer:

1. First, use the substitution u = x−a. This will give us: ∫sinx/sin(x−a) dx = ∫sinx/sinu du

2. Use the formula: ∫sinx/sinu du = -cosx + c

3. Substitute back in for u to get: -cos(x−a) + c

Question:

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

Answer:

1. Rewrite the expression: (1/x^2) * (x^4 + 1)^3/4

2. Use the power rule: (1/x^2) * (3x^12 + 3x^8 + 3x^4 + 1)/4

3. Integrate: (1/x^3) * (x^13/13 + x^9/9 + x^5/5 + x)/4

4. Simplify: (1/x^3) * (x^13 + 3x^9 + 3x^5 + 4x)/52

Question:

Integrate the function 1/sin^3xsin(x+α)

Answer:

1. Integrate 1/sin^3x = -1/2sin^2x + C

2. Substitute sinx = u, du = cosx dx

3. Integrate -1/2u^2du = -1/4u^3 + C

4. Substitute u = sinx, du = cosx dx

5. Integrate -1/4sin^3xcosx dx = -1/4sin^4x + C

6. Integrate -1/4sin^4(x+α) dx = -1/5sin^5(x+α) + C

Question:

Integrate the function √1−√x/1+√x

Answer:

1. Multiply both numerator and denominator by (1+√x)

√1−√x/1+√x * (1+√x)/(1+√x)

2. Simplify the equation

(1−x)/(1+√x)^2

3. Take the integral of both sides

∫(1−x)/(1+√x)^2 dx

4. Use the formula for integration by parts

∫uvdx = uv - ∫vdu

Let u = 1-x and dv = 1/(1+√x)^2 dx

5. Solve for u and v

du = dx v = -2/3(1+√x)^(-3/2)

6. Substitute the values of u and v in the equation

∫(1−x)/(1+√x)^2 dx = -2/3(1+√x)^(-3/2) (1-x) - ∫-2/3(1+√x)^(-3/2)dx

7. Take the integral of both sides

-2/3(1+√x)^(-3/2) (1-x) - (-2/3x(1+√x)^(-3/2) + 2/3(1+√x)^(-1/2))

8. Simplify

2/3(1+√x)^(-1/2) - 2/3x(1+√x)^(-3/2)

Question:

​The other boy asked Ranji to ‘explain’ himself. What did he expect Ranji to say? Was he, in your opinion, right or wrong to ask this question?

Answer:

Step 1: The other boy asked Ranji to explain himself, which means he wanted Ranji to clarify his actions or words.

Step 2: It is difficult to determine whether the other boy was right or wrong to ask this question without knowing the context of the situation. It is possible that the other boy was trying to gain clarity on the situation, or he may have been trying to challenge Ranji’s actions or words.

Step 3: Ultimately, it is up to the reader to decide if the other boy was right or wrong to ask this question as it is a subjective opinion.

Question:

Integrate the function (2+sin2x​/1+cos2x)e^x

Answer:

Answer: Step 1: Rewrite the equation as (2e^x + sin2x e^x)/(1 + cos2x)

Step 2: Multiply the numerator and denominator by 1 + cos2x

Step 3: Rewrite the equation as (2e^x + sin2x e^x + sin2x e^x cos2x)/(1 + cos2x)^2

Step 4: Simplify the numerator by factoring out e^x

Step 5: Rewrite the equation as e^x(2 + sin2x + sin2x cos2x)/(1 + cos2x)^2

Step 6: Use the formula for integration of exponential functions to integrate the equation

Step 7: Integrate (2 + sin2x + sin2x cos2x)/(1 + cos2x)^2

Step 8: The final answer is e^x(2x + sin2x/2 - tan2x/2)/(1 + cos2x)

Question:

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

Answer:

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

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

3. Integrate: -2/3x^-1/3 + 1/4x^4/3 + C

4. Simplify: -2/3x^-1/3 + 1/4x^4/3 + C

Question:

Integrate the function 5x/(x+1)(x^2+9)

Answer:

Step 1: Rewrite the function as a product of two fractions: 5x/(x+1)(x^2+9) = (5x/(x+1))*(1/(x^2+9))

Step 2: Integrate the first fraction: ∫(5x/(x+1))dx = 5ln|x+1| + C

Step 3: Integrate the second fraction: ∫(1/(x^2+9))dx = 1/3*ln|x^2+9| + C

Step 4: Combine the two results: ∫(5x/(x+1)(x^2+9))dx = 5ln|x+1| + 1/3*ln|x^2+9| + C

Question:

Integrate the function : 1/x√ax−x^2

Answer:

1. Separate the function into two parts: 1/x and √ax−x^2
2. Integrate the first part: ln|x| + C
3. Integrate the second part using the substitution u = ax−x^2 u = ax−x^2 du = (a−2x)dx ∫√u du = 2/3u^(3/2) + C
4. Substitute back u = ax−x^2 ∫√ax−x^2 du = 2/3(ax−x^2)^(3/2) + C
5. Combine the two parts: ln|x| + 2/3(ax−x^2)^(3/2) + C

Question:

Integrate the function 1/x−x^3

Answer:

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

Question:

Integrate the function tan^−1√1−x/1+x

Answer:

1. Let u = √1−x/1+x
2. du = -(1+x)^-2 * (1-x)^(1/2) dx
3. Integral tan^-1u du
4. u tan^-1u - ∫ sec^2u du
5. u tan^-1u - ∫ (1 + tan^2u) du
6. u tan^-1u - ∫ (1 + u^2) du
7. u tan^-1u - ∫ u du - ∫ du
8. u tan^-1u - (1/2)u^2 - x + C

Question:

Integrate the function : 1/√x+a+√(x+b)

Answer:

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

Step 2: Integrate the first part: Integral of 1/√x + a = 2√x + ax + C

Step 3: Integrate the second part: Integral of √(x+b) = (2/3)(x+b)^(3/2) + C