IB Mathematics: Analysis and Approaches Quiz: Advanced Integration Techniques
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Advanced Integration TechniquesQuestion 1 of 20

To evaluate the integral ∫cos⁡x4+sin⁡2xdx\int \frac{\cos x}{4 + \sin^2 x} dx, a substitution is made. If the resulting integral is ∫14+u2du\int \frac{1}{4+u^2} du, what was the substitution for uu?

u=cos⁡xu = \cos x
u=sin⁡xu = \sin x
u=sin⁡2xu = \sin^2 x
u=4+sin⁡2xu = 4 + \sin^2 x
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Advanced Integration Techniques

Practice Advanced Integration Techniques in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Advanced Integration Techniques, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

To evaluate the integral ∫cos⁡x4+sin⁡2xdx\int \frac{\cos x}{4 + \sin^2 x} dx, a substitution is made. If the resulting integral is ∫14+u2du\int \frac{1}{4+u^2} du, what was the substitution for uu?

  1. u=cos⁡xu = \cos x
  2. u=sin⁡xu = \sin x (correct answer)
  3. u=sin⁡2xu = \sin^2 x
  4. u=4+sin⁡2xu = 4 + \sin^2 x
Explanation: Let's test the substitution u=sin⁡xu = \sin x. The derivative is dudx=cos⁡x\frac{du}{dx} = \cos x, so du=cos⁡xdxdu = \cos x dx. The denominator becomes 4+(sin⁡x)2=4+u24 + (\sin x)^2 = 4 + u^2. The numerator, cos⁡xdx\cos x dx, becomes dudu. Substituting these into the original integral gives: ∫cos⁡x4+sin⁡2xdx=∫14+u2du\int \frac{\cos x}{4 + \sin^2 x} dx = \int \frac{1}{4+u^2} du This matches the integral given in the question. Therefore, the correct substitution is u=sin⁡xu = \sin x. The other options lead to more complicated or incorrect integrals. For example, if u=cos⁡xu = \cos x, the numerator becomes uu but sin⁡2x=1−u2\sin^2 x = 1-u^2 and dx=−du/sin⁡xdx = -du/\sin x, which does not simplify correctly.

Question 2

The integral I=∫e2xcos⁡x dxI = \int e^{2x} \cos x \, dx is found using integration by parts twice, resulting in an expression of the form e2x(Asin⁡x+Bcos⁡x)+Ce^{2x}(A \sin x + B \cos x) + C.

Determine the value of AA.

  1. −15-\frac{1}{5}
  2. 15\frac{1}{5} (correct answer)
  3. 13\frac{1}{3}
  4. 25\frac{2}{5}
Explanation: We use integration by parts, ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Let u=e2xu = e^{2x} and dv=cos⁡x dxdv = \cos x \, dx. Then du=2e2x dxdu = 2e^{2x} \, dx and v=sin⁡xv = \sin x. I=e2xsin⁡x−∫2e2xsin⁡x dxI = e^{2x} \sin x - \int 2e^{2x} \sin x \, dx Apply integration by parts again to the new integral. Let u=2e2xu = 2e^{2x} and dv=sin⁡x dxdv = \sin x \, dx. Then du=4e2x dxdu = 4e^{2x} \, dx and v=−cos⁡xv = -\cos x. I=e2xsin⁡x−(−2e2xcos⁡x−∫−4e2xcos⁡x dx)I = e^{2x} \sin x - \left( -2e^{2x} \cos x - \int -4e^{2x} \cos x \, dx \right) I=e2xsin⁡x+2e2xcos⁡x−4∫e2xcos⁡x dxI = e^{2x} \sin x + 2e^{2x} \cos x - 4 \int e^{2x} \cos x \, dx Since I=∫e2xcos⁡x dxI = \int e^{2x} \cos x \, dx, we have: I=e2xsin⁡x+2e2xcos⁡x−4II = e^{2x} \sin x + 2e^{2x} \cos x - 4I 5I=e2x(sin⁡x+2cos⁡x)5I = e^{2x}(\sin x + 2\cos x) I=15e2x(sin⁡x+2cos⁡x)+C=e2x(15sin⁡x+25cos⁡x)+CI = \frac{1}{5}e^{2x}(\sin x + 2\cos x) + C = e^{2x}\left(\frac{1}{5}\sin x + \frac{2}{5}\cos x\right) + C Comparing this to e2x(Asin⁡x+Bcos⁡x)+Ce^{2x}(A \sin x + B \cos x) + C, we find A=15A = \frac{1}{5}. Distractor D is the value of BB. Distractor A results from a sign error in the integration by parts formula. Distractor C arises from algebraic mistakes when solving for II.

Question 3

Calculate the exact value of ∫0ln⁡2ex1+e2xdx\int_{0}^{\ln 2} \frac{e^x}{1+e^{2x}} dx.

  1. ln⁡(52)\ln(\frac{5}{2})
  2. arctan⁡(ln⁡2)\arctan(\ln 2)
  3. arctan⁡(2)−π4\arctan(2) - \frac{\pi}{4} (correct answer)
  4. arctan⁡(2)\arctan(2)
Explanation: We use substitution. Let u=exu = e^x. Then du=exdxdu = e^x dx. We must also change the limits of integration. When x=0x=0, u=e0=1u=e^0=1. When x=ln⁡2x=\ln 2, u=eln⁡2=2u=e^{\ln 2}=2. The integral becomes: ∫1211+u2du\int_{1}^{2} \frac{1}{1+u^2} du This is a standard integral: [arctan⁡(u)]12=arctan⁡(2)−arctan⁡(1)\left[\arctan(u)\right]_{1}^{2} = \arctan(2) - \arctan(1) Since arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}, the result is arctan⁡(2)−π4\arctan(2) - \frac{\pi}{4}. Distractor A results from incorrectly integrating 11+u2\frac{1}{1+u^2} to get ln⁡(1+u2)\ln(1+u^2). Distractor B is a common error where the limits of integration are not changed after the substitution. Distractor D results from forgetting to evaluate the integral at the lower limit.

Question 4

Using the substitution u=cos⁡θu = \cos \theta, which of the following integrals is equivalent to ∫sin⁡3θ dθ\int \sin^3 \theta \, d\theta?

  1. ∫(1−u2) du\int (1-u^2) \, du
  2. ∫(u2−1) du\int (u^2-1) \, du (correct answer)
  3. ∫−u2 du\int -u^2 \, du
  4. ∫u1−u2 du\int u \sqrt{1-u^2} \, du
Explanation: To use the substitution u=cos⁡θu = \cos \theta, we first need to express the integrand in terms of cos⁡θ\cos \theta and a single sin⁡θ\sin \theta. We write sin⁡3θ=sin⁡2θ⋅sin⁡θ=(1−cos⁡2θ)sin⁡θ\sin^3 \theta = \sin^2 \theta \cdot \sin \theta = (1-\cos^2 \theta) \sin \theta. So the integral is ∫(1−cos⁡2θ)sin⁡θ dθ\int (1-\cos^2 \theta) \sin \theta \, d\theta. Now, let u=cos⁡θu = \cos \theta. Then du=−sin⁡θ dθdu = -\sin \theta \, d\theta, which means sin⁡θ dθ=−du\sin \theta \, d\theta = -du. Substituting these into the integral gives: ∫(1−u2)(−du)=∫(u2−1) du\int (1-u^2) (-du) = \int (u^2-1) \, du Distractor A is a very common error where the negative sign from du=−sin⁡θ dθdu = -\sin \theta \, d\theta is forgotten. Distractor C arises from incorrectly substituting u2u^2 for sin⁡2θ\sin^2 \theta. Distractor D would arise from an incorrect substitution setup.

Question 5

What is the value of the integral ∫1e2(ln⁡x)2xdx\int_{1}^{e^2} \frac{(\ln x)^2}{x} dx?

  1. 22
  2. 83\frac{8}{3} (correct answer)
  3. 44
  4. e6−13\frac{e^6-1}{3}
Explanation: We use substitution. Let u=ln⁡xu = \ln x. Then du=1xdxdu = \frac{1}{x} dx. We must change the limits of integration. When x=1x=1, u=ln⁡1=0u=\ln 1=0. When x=e2x=e^2, u=ln⁡(e2)=2u=\ln(e^2)=2. The integral becomes: ∫02u2du\int_{0}^{2} u^2 du Now, we integrate with respect to uu: [u33]02=233−033=83\left[\frac{u^3}{3}\right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} Distractor A results from integrating uu instead of u2u^2, giving u2/2u^2/2. Distractor C comes from integrating u2u^2 to u3u^3 but forgetting to divide by 3. Distractor D results from failing to perform the substitution correctly.

Question 6

Find ∫xx−2 dx\int x \sqrt{x-2} \, dx.

  1. 23(x−2)3/2+C\frac{2}{3}(x-2)^{3/2} + C
  2. 25(x−2)5/2+C\frac{2}{5}(x-2)^{5/2} + C
  3. 215(x−2)3/2(3x−16)+C\frac{2}{15}(x-2)^{3/2}(3x-16)+C
  4. 215(x−2)3/2(3x+4)+C\frac{2}{15}(x-2)^{3/2}(3x+4)+C (correct answer)
Explanation: This integral is best solved using substitution. Let u=x−2u = x-2, which means x=u+2x = u+2 and du=dxdu = dx. The integral becomes: ∫(u+2)u du=∫(u3/2+2u1/2) du\int (u+2)\sqrt{u} \, du = \int (u^{3/2} + 2u^{1/2}) \, du Now, integrate with respect to uu: u5/25/2+2u3/23/2+C=25u5/2+43u3/2+C\frac{u^{5/2}}{5/2} + 2\frac{u^{3/2}}{3/2} + C = \frac{2}{5}u^{5/2} + \frac{4}{3}u^{3/2} + C Substitute back u=x−2u=x-2: 25(x−2)5/2+43(x−2)3/2+C\frac{2}{5}(x-2)^{5/2} + \frac{4}{3}(x-2)^{3/2} + C To match the answer choices, factor out 215(x−2)3/2\frac{2}{15}(x-2)^{3/2}: 215(x−2)3/2[3(x−2)+10]+C=215(x−2)3/2(3x−6+10)+C=215(x−2)3/2(3x+4)+C\frac{2}{15}(x-2)^{3/2} [3(x-2) + 10] + C = \frac{2}{15}(x-2)^{3/2} (3x-6+10) + C = \frac{2}{15}(x-2)^{3/2}(3x+4)+C Distractor A ignores the factor of xx in the integrand. Distractor B only integrates the first term that would arise from the substitution, ∫uu du\int u \sqrt{u} \, du. Distractor C comes from an incorrect substitution x=u−2x=u-2 instead of x=u+2x=u+2, leading to a sign error.

Question 7

Which of the following is an expression for ∫sin⁡(ln⁡x) dx\int \sin(\ln x) \, dx?

  1. x(sin⁡(ln⁡x)−cos⁡(ln⁡x))+Cx(\sin(\ln x) - \cos(\ln x)) + C
  2. x2(sin⁡(ln⁡x)+cos⁡(ln⁡x))+C\frac{x}{2}(\sin(\ln x) + \cos(\ln x)) + C
  3. 12(sin⁡(ln⁡x)−cos⁡(ln⁡x))+C\frac{1}{2}(\sin(\ln x) - \cos(\ln x)) + C
  4. x2(sin⁡(ln⁡x)−cos⁡(ln⁡x))+C\frac{x}{2}(\sin(\ln x) - \cos(\ln x)) + C (correct answer)
Explanation: This problem can be solved with integration by parts, or substitution followed by parts. Using parts directly is efficient. Let I=∫sin⁡(ln⁡x)dxI = \int \sin(\ln x) dx. Let u=sin⁡(ln⁡x)u = \sin(\ln x) and dv=dxdv = dx. Then du=cos⁡(ln⁡x)⋅1xdxdu = \cos(\ln x) \cdot \frac{1}{x} dx and v=xv = x. I=xsin⁡(ln⁡x)−∫x⋅cos⁡(ln⁡x)1xdx=xsin⁡(ln⁡x)−∫cos⁡(ln⁡x)dxI = x \sin(\ln x) - \int x \cdot \cos(\ln x) \frac{1}{x} dx = x \sin(\ln x) - \int \cos(\ln x) dx Apply parts to the new integral. Let u=cos⁡(ln⁡x)u = \cos(\ln x) and dv=dxdv = dx. Then du=−sin⁡(ln⁡x)⋅1xdxdu = -\sin(\ln x) \cdot \frac{1}{x} dx and v=xv = x. ∫cos⁡(ln⁡x)dx=xcos⁡(ln⁡x)−∫x(−sin⁡(ln⁡x)1x)dx=xcos⁡(ln⁡x)+∫sin⁡(ln⁡x)dx=xcos⁡(ln⁡x)+I\int \cos(\ln x) dx = x \cos(\ln x) - \int x \left(-\sin(\ln x) \frac{1}{x}\right) dx = x \cos(\ln x) + \int \sin(\ln x) dx = x \cos(\ln x) + I Substitute this back into the equation for II: I=xsin⁡(ln⁡x)−(xcos⁡(ln⁡x)+I)I = x \sin(\ln x) - (x \cos(\ln x) + I) I=xsin⁡(ln⁡x)−xcos⁡(ln⁡x)−II = x \sin(\ln x) - x \cos(\ln x) - I 2I=x(sin⁡(ln⁡x)−cos⁡(ln⁡x))2I = x(\sin(\ln x) - \cos(\ln x)) I=x2(sin⁡(ln⁡x)−cos⁡(ln⁡x))+CI = \frac{x}{2}(\sin(\ln x) - \cos(\ln x)) + C Distractor A results from forgetting to divide by 2 when solving for II. Distractor B has a sign error, corresponding to the result for ∫cos⁡(ln⁡x)dx\int \cos(\ln x) dx. Distractor C misses the factor of xx that arises from the integration.

Question 8

The integral I=∫e2xcos⁡x dxI = \int e^{2x} \cos x \, dx is found using integration by parts twice, resulting in an expression of the form e2x(Asin⁡x+Bcos⁡x)+Ce^{2x}(A \sin x + B \cos x) + C.

Determine the value of AA.

  1. −15-\frac{1}{5}
  2. 15\frac{1}{5} (correct answer)
  3. 13\frac{1}{3}
  4. 25\frac{2}{5}
Explanation: We use integration by parts, ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Let u=e2xu = e^{2x} and dv=cos⁡x dxdv = \cos x \, dx. Then du=2e2x dxdu = 2e^{2x} \, dx and v=sin⁡xv = \sin x. I=e2xsin⁡x−∫2e2xsin⁡x dxI = e^{2x} \sin x - \int 2e^{2x} \sin x \, dx Apply integration by parts again to the new integral. Let u=2e2xu = 2e^{2x} and dv=sin⁡x dxdv = \sin x \, dx. Then du=4e2x dxdu = 4e^{2x} \, dx and v=−cos⁡xv = -\cos x. I=e2xsin⁡x−(−2e2xcos⁡x−∫−4e2xcos⁡x dx)I = e^{2x} \sin x - \left( -2e^{2x} \cos x - \int -4e^{2x} \cos x \, dx \right) I=e2xsin⁡x+2e2xcos⁡x−4∫e2xcos⁡x dxI = e^{2x} \sin x + 2e^{2x} \cos x - 4 \int e^{2x} \cos x \, dx Since I=∫e2xcos⁡x dxI = \int e^{2x} \cos x \, dx, we have: I=e2xsin⁡x+2e2xcos⁡x−4II = e^{2x} \sin x + 2e^{2x} \cos x - 4I 5I=e2x(sin⁡x+2cos⁡x)5I = e^{2x}(\sin x + 2\cos x) I=15e2x(sin⁡x+2cos⁡x)+C=e2x(15sin⁡x+25cos⁡x)+CI = \frac{1}{5}e^{2x}(\sin x + 2\cos x) + C = e^{2x}\left(\frac{1}{5}\sin x + \frac{2}{5}\cos x\right) + C Comparing this to e2x(Asin⁡x+Bcos⁡x)+Ce^{2x}(A \sin x + B \cos x) + C, we find A=15A = \frac{1}{5}. Distractor D is the value of BB. Distractor A results from a sign error in the integration by parts formula. Distractor C arises from algebraic mistakes when solving for II.

Question 9

The value of ∫0π/2xcos⁡(x)dx\int_{0}^{\pi/2} x \cos(x) dx is:

  1. 11
  2. π2\frac{\pi}{2}
  3. π2−1\frac{\pi}{2} - 1 (correct answer)
  4. π2+1\frac{\pi}{2} + 1
Explanation: We use integration by parts. Let u=xu=x and dv=cos⁡(x)dxdv = \cos(x) dx. Then du=dxdu=dx and v=sin⁡(x)v=\sin(x). ∫0π/2xcos⁡(x)dx=[xsin⁡(x)]0π/2−∫0π/2sin⁡(x)dx\int_{0}^{\pi/2} x \cos(x) dx = [x \sin(x)]_{0}^{\pi/2} - \int_{0}^{\pi/2} \sin(x) dx =[xsin⁡(x)−(−cos⁡(x))]0π/2=[xsin⁡(x)+cos⁡(x)]0π/2= [x \sin(x) - (-\cos(x))]_{0}^{\pi/2} = [x \sin(x) + \cos(x)]_{0}^{\pi/2} Now, evaluate at the limits: (π2sin⁡(π2)+cos⁡(π2))−(0⋅sin⁡(0)+cos⁡(0))\left(\frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right) - (0 \cdot \sin(0) + \cos(0)) =(π2⋅1+0)−(0+1)= \left(\frac{\pi}{2} \cdot 1 + 0\right) - (0 + 1) =π2−1= \frac{\pi}{2} - 1 Distractor A results from calculation errors, possibly cos⁡(0)=0\cos(0)=0. Distractor B results from ignoring the second term ∫vdu\int v du. Distractor D comes from a sign error when integrating sin⁡(x)\sin(x), i.e., ∫sin⁡(x)dx=cos⁡(x)\int \sin(x) dx = \cos(x).

Question 10

Let g(x)=∫1x2sin⁡(t)tdtg(x) = \int_{1}^{x^2} \frac{\sin(t)}{\sqrt{t}} dt. What is the value of g′(π)g'(\sqrt{\pi})?

  1. −2π-\frac{2}{\sqrt{\pi}}
  2. 00 (correct answer)
  3. 11
  4. 22
Explanation: This question requires the application of the Fundamental Theorem of Calculus Part 2, combined with the chain rule. If F(x)=∫ah(x)f(t)dtF(x) = \int_{a}^{h(x)} f(t) dt, then F′(x)=f(h(x))⋅h′(x)F'(x) = f(h(x)) \cdot h'(x). In this case, f(t)=sin⁡(t)tf(t) = \frac{\sin(t)}{\sqrt{t}} and the upper limit is h(x)=x2h(x) = x^2. The derivative of the upper limit is h′(x)=2xh'(x) = 2x. So, g′(x)=sin⁡(x2)x2⋅(2x)g'(x) = \frac{\sin(x^2)}{\sqrt{x^2}} \cdot (2x). For x>0x>0, x2=x\sqrt{x^2}=x, so g′(x)=sin⁡(x2)x⋅(2x)=2sin⁡(x2)g'(x) = \frac{\sin(x^2)}{x} \cdot (2x) = 2\sin(x^2). We need to find g′(π)g'(\sqrt{\pi}): g′(π)=2sin⁡((π)2)=2sin⁡(π)=2⋅0=0g'(\sqrt{\pi}) = 2\sin((\sqrt{\pi})^2) = 2\sin(\pi) = 2 \cdot 0 = 0 Distractor A results from miscalculation or errors in applying the chain rule. Distractor D might arise from forgetting the sin⁡\sin function and evaluating 22 at x=πx=\sqrt{\pi}.

Question 11

Find the exact value of ∫01xe2xdx\int_{0}^{1} x e^{2x} dx.

  1. e2−14\frac{e^2-1}{4}
  2. e2+14\frac{e^2+1}{4} (correct answer)
  3. e22\frac{e^2}{2}
  4. e2+12\frac{e^2+1}{2}
Explanation: This integral requires integration by parts, using the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Let u=xu = x and dv=e2xdxdv = e^{2x} dx. Then du=dxdu = dx and v=12e2xv = \frac{1}{2}e^{2x}. Applying the formula for the definite integral: ∫01xe2xdx=[x⋅12e2x]01−∫0112e2xdx\int_{0}^{1} x e^{2x} dx = \left[x \cdot \frac{1}{2}e^{2x}\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}e^{2x} dx =[12xe2x]01−[14e2x]01= \left[\frac{1}{2}xe^{2x}\right]_{0}^{1} - \left[\frac{1}{4}e^{2x}\right]_{0}^{1} =(12(1)e2−12(0)e0)−(14e2−14e0)= \left(\frac{1}{2}(1)e^{2} - \frac{1}{2}(0)e^{0}\right) - \left(\frac{1}{4}e^{2} - \frac{1}{4}e^{0}\right) =12e2−(14e2−14)= \frac{1}{2}e^{2} - \left(\frac{1}{4}e^{2} - \frac{1}{4}\right) =12e2−14e2+14=14e2+14=e2+14= \frac{1}{2}e^{2} - \frac{1}{4}e^{2} + \frac{1}{4} = \frac{1}{4}e^{2} + \frac{1}{4} = \frac{e^2+1}{4} Distractor A results from a sign error in the formula, calculating uv−(−∫v du)uv - (-\int v\,du) which becomes uv+∫v duuv+\int v\,du but then making another sign error in evaluation. Distractor C occurs if one forgets that e0=1e^0=1 and treats the lower bound evaluation as zero. Distractor D results from forgetting the factor of 1/21/2 when integrating e2xe^{2x} the first time, i.e., incorrectly taking v=e2xv=e^{2x}.

Question 12

Evaluate ∫x2sin⁡x dx\int x^2 \sin x \, dx.

  1. −x2cos⁡x−2xsin⁡x+2cos⁡x+C-x^2 \cos x - 2x \sin x + 2\cos x + C
  2. x2cos⁡x−2xsin⁡x−2cos⁡x+Cx^2 \cos x - 2x \sin x - 2\cos x + C
  3. −x2cos⁡x+2xsin⁡x+2cos⁡x+C-x^2 \cos x + 2x \sin x + 2\cos x + C (correct answer)
  4. −x2cos⁡x+2xsin⁡x−2cos⁡x+C-x^2 \cos x + 2x \sin x - 2\cos x + C
Explanation: This requires repeated integration by parts. Let's use the tabular method for efficiency. Let uu-column be for differentiation and dvdv-column for integration.
DifferentiateIntegrate
x2x^2sin⁡x\sin x
2x2x−cos⁡x-\cos x
22−sin⁡x-\sin x
00cos⁡x\cos x
Multiply diagonally with alternating signs (+, -, +):
  1. +(x2)(−cos⁡x)=−x2cos⁡x+(x^2)(-\cos x) = -x^2 \cos x
  2. −(2x)(−sin⁡x)=+2xsin⁡x-(2x)(-\sin x) = +2x \sin x
  3. +(2)(cos⁡x)=+2cos⁡x+(2)(\cos x) = +2\cos x
Summing these terms gives the integral: ∫x2sin⁡x dx=−x2cos⁡x+2xsin⁡x+2cos⁡x+C\int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2\cos x + C Distractors A, B, and D are variations with incorrect signs, which commonly arise from errors in the alternating signs of the tabular method or in the integrals/derivatives of sin and cos.

Question 13

Find ∫arctan⁡(x)dx\int \arctan(x) dx.

  1. xarctan⁡(x)−ln⁡(1+x2)+Cx \arctan(x) - \ln(1+x^2) + C
  2. xarctan⁡(x)+12ln⁡(1+x2)+Cx \arctan(x) + \frac{1}{2} \ln(1+x^2) + C
  3. xarctan⁡(x)+1−x2+Cx \arctan(x) + \sqrt{1-x^2} + C
  4. xarctan⁡(x)−12ln⁡(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1+x^2) + C (correct answer)
Explanation: This is a standard integration by parts problem where dv=dxdv=dx. Let u=arctan⁡(x)u = \arctan(x) and dv=dxdv = dx. Then du=11+x2dxdu = \frac{1}{1+x^2} dx and v=xv = x. Using the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: ∫arctan⁡(x)dx=xarctan⁡(x)−∫x⋅11+x2dx=xarctan⁡(x)−∫x1+x2dx\int \arctan(x) dx = x \arctan(x) - \int x \cdot \frac{1}{1+x^2} dx = x \arctan(x) - \int \frac{x}{1+x^2} dx To evaluate the remaining integral, we use substitution. Let w=1+x2w = 1+x^2, so dw=2xdxdw = 2x dx, or xdx=12dwx dx = \frac{1}{2} dw. ∫x1+x2dx=∫1w(12dw)=12ln⁡∣w∣+C=12ln⁡(1+x2)+C\int \frac{x}{1+x^2} dx = \int \frac{1}{w} \left(\frac{1}{2} dw\right) = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C (Absolute value is not needed as 1+x2>01+x^2 > 0). Combining the results gives: xarctan⁡(x)−12ln⁡(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1+x^2) + C Distractor A results from forgetting the factor of 1/21/2 from the substitution. Distractor B has a sign error from the integration by parts formula. Distractor C arises from incorrectly remembering the derivative of arctan⁡(x)\arctan(x) as the derivative of arcsin⁡(x)\arcsin(x).

Question 14

Determine the value of ∫1eln⁡xx2dx\int_{1}^{e} \frac{\ln x}{x^2} dx.

  1. −1-1
  2. 2e−1\frac{2}{e} - 1
  3. 1−2e1 - \frac{2}{e} (correct answer)
  4. 2−2e2 - \frac{2}{e}
Explanation: We use integration by parts. Let u=ln⁡xu = \ln x and dv=1x2dx=x−2dxdv = \frac{1}{x^2} dx = x^{-2} dx. Then du=1xdxdu = \frac{1}{x} dx and v=−x−1=−1xv = -x^{-1} = -\frac{1}{x}. Applying the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: ∫1eln⁡xx2dx=[−ln⁡xx]1e−∫1e(−1x)1xdx\int_{1}^{e} \frac{\ln x}{x^2} dx = \left[-\frac{\ln x}{x}\right]_{1}^{e} - \int_{1}^{e} \left(-\frac{1}{x}\right) \frac{1}{x} dx =[−ln⁡xx]1e+∫1e1x2dx= \left[-\frac{\ln x}{x}\right]_{1}^{e} + \int_{1}^{e} \frac{1}{x^2} dx =[−ln⁡xx−1x]1e= \left[-\frac{\ln x}{x} - \frac{1}{x}\right]_{1}^{e} Now evaluate at the limits: =(−ln⁡ee−1e)−(−ln⁡11−11)= \left(-\frac{\ln e}{e} - \frac{1}{e}\right) - \left(-\frac{\ln 1}{1} - \frac{1}{1}\right) =(−1e−1e)−(0−1)= \left(-\frac{1}{e} - \frac{1}{e}\right) - \left(0 - 1\right) =−2e−(−1)=1−2e= -\frac{2}{e} - (-1) = 1 - \frac{2}{e} Distractor B comes from an error in the sign when integrating dvdv, taking v=1/xv = 1/x. Distractor D results from incorrectly evaluating ln⁡1=1\ln 1 = 1 instead of 0. Distractor A arises from a combination of sign errors.

Question 15

Given that ∫a2a1xln⁡xdx=ln⁡3\int_{a}^{2a} \frac{1}{x \ln x} dx = \ln 3 for a>1a>1, find the value of aa.

  1. 2\sqrt{2} (correct answer)
  2. 3\sqrt{3}
  3. 22
  4. 33
Explanation: To solve the integral ∫1xln⁡xdx\int \frac{1}{x \ln x} dx, we use the substitution u=ln⁡xu = \ln x. This gives du=1xdxdu = \frac{1}{x} dx. We must also change the limits of integration. When x=ax=a, u=ln⁡au=\ln a. When x=2ax=2a, u=ln⁡(2a)u=\ln(2a). The integral becomes: ∫ln⁡aln⁡(2a)1udu=[ln⁡∣u∣]ln⁡aln⁡(2a)\int_{\ln a}^{\ln(2a)} \frac{1}{u} du = [\ln|u|]_{\ln a}^{\ln(2a)} Since a>1a>1, ln⁡a>0\ln a > 0 and ln⁡(2a)>0\ln(2a) > 0. ln⁡(ln⁡(2a))−ln⁡(ln⁡a)=ln⁡(ln⁡(2a)ln⁡a)\ln(\ln(2a)) - \ln(\ln a) = \ln\left(\frac{\ln(2a)}{\ln a}\right) We are given that this value is ln⁡3\ln 3. Therefore: ln⁡(2a)ln⁡a=3\frac{\ln(2a)}{\ln a} = 3 ln⁡(2a)=3ln⁡a\ln(2a) = 3 \ln a Using logarithm properties, 3ln⁡a=ln⁡(a3)3 \ln a = \ln(a^3). ln⁡(2a)=ln⁡(a3)\ln(2a) = \ln(a^3) 2a=a32a = a^3 Since a>1a>1, we can divide by aa: 2=a2  ⟹  a=22 = a^2 \implies a = \sqrt{2} Distractor C, a=2a=2, might result from the error ln⁡(2a)=ln⁡(2)+ln⁡(a)\ln(2a) = \ln(2) + \ln(a), leading to ln⁡(2)+ln⁡(a)=3ln⁡(a)\ln(2)+\ln(a) = 3\ln(a) or ln⁡(2)=2ln⁡(a)=ln⁡(a2)\ln(2) = 2\ln(a) = \ln(a^2), so a2=2a^2=2. This logic is correct. What error leads to a=2a=2? Perhaps ln⁡(2a)ln⁡a=ln⁡(2)\frac{\ln(2a)}{\ln a} = \ln(2)? No. Perhaps 2aa=3\frac{2a}{a}=3? No. An error in log rules like ln⁡(2a)=2ln⁡(a)\ln(2a) = 2\ln(a) might lead to 2ln⁡(a)=3ln⁡(a)2\ln(a)=3\ln(a), which means ln⁡(a)=0\ln(a)=0, so a=1a=1, not an option. Distractor B may come from setting the expression equal to 3 instead of ln⁡3\ln 3. Distractor C, a=2, is a common guess.

Question 16

Given that ∫a2a1xln⁡xdx=ln⁡3\int_{a}^{2a} \frac{1}{x \ln x} dx = \ln 3 for a>1a>1, find the value of aa.

  1. 2\sqrt{2} (correct answer)
  2. 3\sqrt{3}
  3. 22
  4. 33
Explanation: To solve the integral ∫1xln⁡xdx\int \frac{1}{x \ln x} dx, we use the substitution u=ln⁡xu = \ln x. This gives du=1xdxdu = \frac{1}{x} dx. We must also change the limits of integration. When x=ax=a, u=ln⁡au=\ln a. When x=2ax=2a, u=ln⁡(2a)u=\ln(2a). The integral becomes: ∫ln⁡aln⁡(2a)1udu=[ln⁡∣u∣]ln⁡aln⁡(2a)\int_{\ln a}^{\ln(2a)} \frac{1}{u} du = [\ln|u|]_{\ln a}^{\ln(2a)} Since a>1a>1, ln⁡a>0\ln a > 0 and ln⁡(2a)>0\ln(2a) > 0. ln⁡(ln⁡(2a))−ln⁡(ln⁡a)=ln⁡(ln⁡(2a)ln⁡a)\ln(\ln(2a)) - \ln(\ln a) = \ln\left(\frac{\ln(2a)}{\ln a}\right) We are given that this value is ln⁡3\ln 3. Therefore: ln⁡(2a)ln⁡a=3\frac{\ln(2a)}{\ln a} = 3 ln⁡(2a)=3ln⁡a\ln(2a) = 3 \ln a Using logarithm properties, 3ln⁡a=ln⁡(a3)3 \ln a = \ln(a^3). ln⁡(2a)=ln⁡(a3)\ln(2a) = \ln(a^3) 2a=a32a = a^3 Since a>1a>1, we can divide by aa: 2=a2  ⟹  a=22 = a^2 \implies a = \sqrt{2} Distractor C, a=2a=2, might result from the error ln⁡(2a)=ln⁡(2)+ln⁡(a)\ln(2a) = \ln(2) + \ln(a), leading to ln⁡(2)+ln⁡(a)=3ln⁡(a)\ln(2)+\ln(a) = 3\ln(a) or ln⁡(2)=2ln⁡(a)=ln⁡(a2)\ln(2) = 2\ln(a) = \ln(a^2), so a2=2a^2=2. This logic is correct. What error leads to a=2a=2? Perhaps ln⁡(2a)ln⁡a=ln⁡(2)\frac{\ln(2a)}{\ln a} = \ln(2)? No. Perhaps 2aa=3\frac{2a}{a}=3? No. An error in log rules like ln⁡(2a)=2ln⁡(a)\ln(2a) = 2\ln(a) might lead to 2ln⁡(a)=3ln⁡(a)2\ln(a)=3\ln(a), which means ln⁡(a)=0\ln(a)=0, so a=1a=1, not an option. Distractor B may come from setting the expression equal to 3 instead of ln⁡3\ln 3. Distractor C, a=2, is a common guess.

Question 17

Find ∫arctan⁡(x)dx\int \arctan(x) dx.

  1. xarctan⁡(x)−ln⁡(1+x2)+Cx \arctan(x) - \ln(1+x^2) + C
  2. xarctan⁡(x)+12ln⁡(1+x2)+Cx \arctan(x) + \frac{1}{2} \ln(1+x^2) + C
  3. xarctan⁡(x)+1−x2+Cx \arctan(x) + \sqrt{1-x^2} + C
  4. xarctan⁡(x)−12ln⁡(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1+x^2) + C (correct answer)
Explanation: This is a standard integration by parts problem where dv=dxdv=dx. Let u=arctan⁡(x)u = \arctan(x) and dv=dxdv = dx. Then du=11+x2dxdu = \frac{1}{1+x^2} dx and v=xv = x. Using the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: ∫arctan⁡(x)dx=xarctan⁡(x)−∫x⋅11+x2dx=xarctan⁡(x)−∫x1+x2dx\int \arctan(x) dx = x \arctan(x) - \int x \cdot \frac{1}{1+x^2} dx = x \arctan(x) - \int \frac{x}{1+x^2} dx To evaluate the remaining integral, we use substitution. Let w=1+x2w = 1+x^2, so dw=2xdxdw = 2x dx, or xdx=12dwx dx = \frac{1}{2} dw. ∫x1+x2dx=∫1w(12dw)=12ln⁡∣w∣+C=12ln⁡(1+x2)+C\int \frac{x}{1+x^2} dx = \int \frac{1}{w} \left(\frac{1}{2} dw\right) = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C (Absolute value is not needed as 1+x2>01+x^2 > 0). Combining the results gives: xarctan⁡(x)−12ln⁡(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1+x^2) + C Distractor A results from forgetting the factor of 1/21/2 from the substitution. Distractor B has a sign error from the integration by parts formula. Distractor C arises from incorrectly remembering the derivative of arctan⁡(x)\arctan(x) as the derivative of arcsin⁡(x)\arcsin(x).

Question 18

Determine the value of ∫1eln⁡xx2dx\int_{1}^{e} \frac{\ln x}{x^2} dx.

  1. −1-1
  2. 2e−1\frac{2}{e} - 1
  3. 1−2e1 - \frac{2}{e} (correct answer)
  4. 2−2e2 - \frac{2}{e}
Explanation: We use integration by parts. Let u=ln⁡xu = \ln x and dv=1x2dx=x−2dxdv = \frac{1}{x^2} dx = x^{-2} dx. Then du=1xdxdu = \frac{1}{x} dx and v=−x−1=−1xv = -x^{-1} = -\frac{1}{x}. Applying the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: ∫1eln⁡xx2dx=[−ln⁡xx]1e−∫1e(−1x)1xdx\int_{1}^{e} \frac{\ln x}{x^2} dx = \left[-\frac{\ln x}{x}\right]_{1}^{e} - \int_{1}^{e} \left(-\frac{1}{x}\right) \frac{1}{x} dx =[−ln⁡xx]1e+∫1e1x2dx= \left[-\frac{\ln x}{x}\right]_{1}^{e} + \int_{1}^{e} \frac{1}{x^2} dx =[−ln⁡xx−1x]1e= \left[-\frac{\ln x}{x} - \frac{1}{x}\right]_{1}^{e} Now evaluate at the limits: =(−ln⁡ee−1e)−(−ln⁡11−11)= \left(-\frac{\ln e}{e} - \frac{1}{e}\right) - \left(-\frac{\ln 1}{1} - \frac{1}{1}\right) =(−1e−1e)−(0−1)= \left(-\frac{1}{e} - \frac{1}{e}\right) - \left(0 - 1\right) =−2e−(−1)=1−2e= -\frac{2}{e} - (-1) = 1 - \frac{2}{e} Distractor B comes from an error in the sign when integrating dvdv, taking v=1/xv = 1/x. Distractor D results from incorrectly evaluating ln⁡1=1\ln 1 = 1 instead of 0. Distractor A arises from a combination of sign errors.

Question 19

Which substitution would be most effective to find the integral ∫1ex+e−xdx\int \frac{1}{e^x + e^{-x}} dx?

  1. u=ex+e−xu = e^x + e^{-x}
  2. u=exu = e^x (correct answer)
  3. u=e−xu = e^{-x}
  4. u=xu = x
Explanation: First, algebraically manipulate the integrand: ∫1ex+e−xdx=∫1ex+1exdx=∫1e2x+1exdx=∫exe2x+1dx\int \frac{1}{e^x + e^{-x}} dx = \int \frac{1}{e^x + \frac{1}{e^x}} dx = \int \frac{1}{\frac{e^{2x}+1}{e^x}} dx = \int \frac{e^x}{e^{2x}+1} dx Now, the integral is in a form suitable for substitution. Let u=exu = e^x. Then du=exdxdu = e^x dx and e2x=(ex)2=u2e^{2x} = (e^x)^2 = u^2. The integral becomes: ∫1u2+1du\int \frac{1}{u^2+1} du This is a standard integral (arctan⁡(u)+C\arctan(u) + C), so the substitution u=exu=e^x is effective. Distractor A, u=ex+e−xu = e^x + e^{-x}, leads to du=(ex−e−x)dxdu = (e^x - e^{-x})dx, which does not simplify the integral. Distractor C, u=e−xu=e^{-x}, is also workable but u=exu=e^x is more direct. Distractor D is not a substitution.

Question 20

Let f(x)f(x) be a differentiable function. Which of the following is equivalent to ∫xf′′(x)dx\int x f''(x) dx?

  1. xf′(x)−f(x)+Cx f'(x) - f(x) + C (correct answer)
  2. xf′(x)+f(x)+Cx f'(x) + f(x) + C
  3. x22f′′(x)−xf′(x)+C\frac{x^2}{2} f''(x) - x f'(x) + C
  4. xf(x)−∫f(x)dx+Cx f(x) - \int f(x) dx + C
Explanation: We use integration by parts with u=xu=x and dv=f′′(x)dxdv=f''(x)dx. From these choices, we have: du=dxdu = dx v=∫f′′(x)dx=f′(x)v = \int f''(x)dx = f'(x) Applying the integration by parts formula, ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du: ∫xf′′(x)dx=xf′(x)−∫f′(x)dx\int x f''(x) dx = x f'(x) - \int f'(x) dx Since ∫f′(x)dx=f(x)+C1\int f'(x) dx = f(x) + C_1, the expression becomes: xf′(x)−(f(x)+C1)=xf′(x)−f(x)+Cx f'(x) - (f(x) + C_1) = x f'(x) - f(x) + C where C=−C1C = -C_1 is the constant of integration. Distractor B comes from a sign error in the integration by parts formula (uv+∫v duuv + \int v\,du). Distractor C results from an incorrect application of integration by parts. Distractor D arises from incorrectly integrating f′′(x)f''(x) to get f(x)f(x).