Calculus 2 Quiz: Integration By Parts
20 questions · exam conditions
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Integration By PartsQuestion 1 of 20

Applying integration by parts to the integral ∫ln⁡(1−x)x dx\int \frac{\ln(1-x)}{x} \,dx with u=ln⁡(1−x)u = \ln(1-x) and dv=1x dxdv = \frac{1}{x} \,dx results in which expression?

ln⁡(x)ln⁡(1−x)−∫ln⁡(x)1−x dx\ln(x)\ln(1-x) - \int \frac{\ln(x)}{1-x} \,dx
ln⁡(x)ln⁡(1−x)+∫ln⁡(x)1−x dx\ln(x)\ln(1-x) + \int \frac{\ln(x)}{1-x} \,dx
(ln⁡(1−x))22−∫ln⁡(x) dx\frac{(\ln(1-x))^2}{2} - \int \ln(x) \,dx
xln⁡(1−x)−∫x1−x dxx\ln(1-x) - \int \frac{x}{1-x} \,dx
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Calculus 2 Quiz

Calculus 2 Quiz: Integration By Parts

Practice Integration By Parts in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integration By Parts, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Applying integration by parts to the integral ∫ln⁡(1−x)x dx\int \frac{\ln(1-x)}{x} \,dx with u=ln⁡(1−x)u = \ln(1-x) and dv=1x dxdv = \frac{1}{x} \,dx results in which expression?

  1. ln⁡(x)ln⁡(1−x)−∫ln⁡(x)1−x dx\ln(x)\ln(1-x) - \int \frac{\ln(x)}{1-x} \,dx
  2. ln⁡(x)ln⁡(1−x)+∫ln⁡(x)1−x dx\ln(x)\ln(1-x) + \int \frac{\ln(x)}{1-x} \,dx (correct answer)
  3. (ln⁡(1−x))22−∫ln⁡(x) dx\frac{(\ln(1-x))^2}{2} - \int \ln(x) \,dx
  4. xln⁡(1−x)−∫x1−x dxx\ln(1-x) - \int \frac{x}{1-x} \,dx
Explanation: Given the choice u=ln⁡(1−x)u = \ln(1-x) and dv=1x dxdv = \frac{1}{x} \,dx. We find the differentials and integrals: du=−11−x dxdu = \frac{-1}{1-x} \,dx and v=ln⁡(x)v = \ln(x). Using the integration by parts formula ∫u dv=uv−∫v du\int u \,dv = uv - \int v \,du, we get: ∫ln⁡(1−x)x dx=ln⁡(1−x)ln⁡(x)−∫ln⁡(x)(−11−x) dx\int \frac{\ln(1-x)}{x} \,dx = \ln(1-x)\ln(x) - \int \ln(x) \left( \frac{-1}{1-x} \right) \,dx. Simplifying the expression gives ln⁡(x)ln⁡(1−x)+∫ln⁡(x)1−x dx\ln(x)\ln(1-x) + \int \frac{\ln(x)}{1-x} \,dx.

Question 2

Evaluate the definite integral ∫1e2ln⁡(x)x dx\int_{1}^{e^2} \frac{\ln(x)}{\sqrt{x}} \,dx.

  1. 4e−44e - 4
  2. 4e4e
  3. 44 (correct answer)
  4. 22
Explanation: Use integration by parts with u=ln⁡(x)u = \ln(x) and dv=x−1/2dxdv = x^{-1/2}dx. Then du=1xdxdu = \frac{1}{x}dx and v=2x1/2=2xv = 2x^{1/2} = 2\sqrt{x}. The integral is \[2\sqrt{x}\ln(x)\]_{1}^{e^2} - \int_{1}^{e^2} 2\sqrt{x} \cdot \frac{1}{x} \,dx. Evaluate the first term: 2e2ln⁡(e2)−21ln⁡(1)=2e(2)−2(1)(0)=4e2\sqrt{e^2}\ln(e^2) - 2\sqrt{1}\ln(1) = 2e(2) - 2(1)(0) = 4e. The integral term is −∫1e22x−1/2 dx=−[4x1/2]1e2=−[4x]1e2=−(4e2−41)=−(4e−4)- \int_{1}^{e^2} 2x^{-1/2} \,dx = -[4x^{1/2}]_{1}^{e^2} = -[4\sqrt{x}]_{1}^{e^2} = -(4\sqrt{e^2} - 4\sqrt{1}) = -(4e - 4). Combining the parts gives 4e−(4e−4)=44e - (4e - 4) = 4.

Question 3

Evaluate the definite integral ∫01(x2+1)e−x dx\int_{0}^{1} (x^2+1)e^{-x} \,dx.

  1. 3−6/e3 - 6/e (correct answer)
  2. 3−4/e3 - 4/e
  3. 1−2/e1 - 2/e
  4. 2−5/e2 - 5/e
Explanation: We can use tabular integration or two applications of integration by parts. Let's use the tabular method with u=x2+1u = x^2+1 and dv=e−xdxdv = e^{-x} dx. The derivatives of uu are 2x2x, 22, and 00. The integrals of dvdv are −e−x-e^{-x}, e−xe^{-x}, and −e−x-e^{-x}. The antiderivative is (x2+1)(−e−x)−(2x)(e−x)+(2)(−e−x)=−e−x(x2+1+2x+2)=−e−x(x2+2x+3)(x^2+1)(-e^{-x}) - (2x)(e^{-x}) + (2)(-e^{-x}) = -e^{-x}(x^2+1+2x+2) = -e^{-x}(x^2+2x+3). Now we evaluate this from 0 to 1: [−e−x(x2+2x+3)]01=(−e−1(12+2(1)+3))−(−e0(02+2(0)+3))=−e−1(6)−(−1)(3)=3−6/e[-e^{-x}(x^2+2x+3)]_{0}^{1} = (-e^{-1}(1^2+2(1)+3)) - (-e^{0}(0^2+2(0)+3)) = -e^{-1}(6) - (-1)(3) = 3 - 6/e.

Question 4

The integral ∫arctan⁡xdx\int \arctan x dx can be evaluated using integration by parts. If the result is xarctan⁡x−12ln⁡(1+x2)+Cx \arctan x - \frac{1}{2} \ln(1 + x^2) + C, which choice for uu and dvdv was used?

  1. u=1,dv=arctan⁡xdxu = 1, dv = \arctan x dx
  2. u=x,dv=arctan⁡xdxu = x, dv = \arctan x dx
  3. u=11+x2,dv=xdxu = \frac{1}{1+x^2}, dv = x dx
  4. u=arctan⁡x,dv=dxu = \arctan x, dv = dx (correct answer)
Explanation: When you encounter an integral like ∫arctan⁡x dx\int \arctan x \, dx, integration by parts is the natural choice since inverse trigonometric functions are typically hard to integrate directly but easy to differentiate. The integration by parts formula is ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. The key strategy is choosing uu to be the function that becomes simpler when differentiated, and dvdv to be something you can easily integrate. For ∫arctan⁡x dx\int \arctan x \, dx, the correct choice is u=arctan⁡xu = \arctan x and dv=dxdv = dx. This gives us du=11+x2dxdu = \frac{1}{1+x^2} dx and v=xv = x. Applying the formula: ∫arctan⁡x dx=xarctan⁡x−∫x1+x2dx\int \arctan x \, dx = x \arctan x - \int \frac{x}{1+x^2} dx. The remaining integral evaluates to 12ln⁡(1+x2)\frac{1}{2}\ln(1+x^2), giving the final result xarctan⁡x−12ln⁡(1+x2)+Cx \arctan x - \frac{1}{2}\ln(1+x^2) + C. This confirms choice D is correct. Choice A (u=1,dv=arctan⁡x dxu = 1, dv = \arctan x \, dx) fails because you can't easily integrate arctan⁡x\arctan x to find vv. Choice B (u=x,dv=arctan⁡x dxu = x, dv = \arctan x \, dx) has the same problem—you'd need v=∫arctan⁡x dxv = \int \arctan x \, dx, which is the original integral you're trying to solve. Choice C (u=11+x2,dv=x dxu = \frac{1}{1+x^2}, dv = x \, dx) doesn't match our original integral ∫arctan⁡x dx\int \arctan x \, dx at all. Remember: for integration by parts with inverse trig functions, set the inverse trig function as uu and the remaining part as dvdv.

Question 5

To evaluate ∫x2e−xdx\int x^2 e^{-x} dx using integration by parts, what is the most efficient sequence of choices for uu and dvdv?

  1. First: u=x2,dv=e−xdxu = x^2, dv = e^{-x}dx; Second: u=x,dv=e−xdxu = x, dv = e^{-x}dx (correct answer)
  2. First: u=e−x,dv=x2dxu = e^{-x}, dv = x^2 dx; Second: u=e−x,dv=xdxu = e^{-x}, dv = x dx
  3. First: u=x,dv=xe−xdxu = x, dv = xe^{-x}dx; Second: u=x,dv=e−xdxu = x, dv = e^{-x}dx
  4. First: u=x2,dv=e−xdxu = x^2, dv = e^{-x}dx; Second: u=xe−x,dv=dxu = xe^{-x}, dv = dx
Explanation: For polynomials multiplied by exponentials, choose the polynomial as u (to reduce its degree through differentiation) and the exponential as dv (since it's easy to integrate). First application: u = x², dv = e^(-x)dx gives du = 2x dx, v = -e^(-x). This reduces the degree from x² to x. Second application on the remaining ∫x e^(-x)dx: u = x, dv = e^(-x)dx gives du = dx, v = -e^(-x). Choice B would make integration increasingly difficult. Choice C has an incorrect dv in the first step. Choice D incorrectly treats xe^(-x) as dv in the second step.

Question 6

When using integration by parts to evaluate an integral of the form ∫p(x)f(x) dx\int p(x) f(x) \,dx where p(x)p(x) is a polynomial, which condition makes the choice u=p(x)u=p(x) and dv=f(x)dxdv=f(x)dx most effective?

  1. The integral of p(x)p(x) is simpler than p(x)p(x), and the derivative of f(x)f(x) is easy to find.
  2. Both p(x)p(x) and f(x)f(x) can be easily integrated and differentiated.
  3. The derivative of f(x)f(x) is simpler than f(x)f(x), and the integral of p(x)p(x) is available.
  4. The derivative of p(x)p(x) is simpler than p(x)p(x), and the integral of f(x)f(x) is available. (correct answer)
Explanation: The formula for integration by parts is ∫u dv=uv−∫v du\int u \,dv = uv - \int v \,du. The strategy is to choose uu and dvdv such that the new integral, ∫v du\int v \,du, is simpler to solve than the original. When p(x)p(x) is a polynomial, its derivatives become progressively simpler, eventually becoming zero. If we choose u=p(x)u=p(x), then du=p′(x)dxdu = p'(x)dx is a polynomial of a lower degree, making the new integral simpler. This strategy is only viable if we can find v=∫dv=∫f(x)dxv = \int dv = \int f(x)dx. Therefore, the choice is effective if the derivative of p(x)p(x) is simpler and the integral of f(x)f(x) is known.

Question 7

When using the tabular method for integration by parts on an integral ∫p(x)f(x) dx\int p(x)f(x) \,dx, where p(x)p(x) is a polynomial, the process terminates. This termination occurs because:

  1. The integrals of f(x)f(x) eventually become zero.
  2. The product of the derivatives of p(x)p(x) and integrals of f(x)f(x) eventually becomes a constant.
  3. The derivatives of the polynomial p(x)p(x) eventually become zero. (correct answer)
  4. The process becomes cyclic, allowing the integral to be solved algebraically.
Explanation: The tabular method is a streamlined version of repeated integration by parts. One column consists of the function chosen as uu (in this case, the polynomial p(x)p(x)) and its successive derivatives. The other column consists of the function chosen as dvdv (in this case, f(x)dxf(x)dx) and its successive integrals. Since p(x)p(x) is a polynomial of degree nn, its (n+1)(n+1)-th derivative will be zero. This causes the process to terminate, as the terms in the expansion will eventually become zero.

Question 8

Consider the integral ∫x2x+1dx\int x^2 \sqrt{x+1} dx. To evaluate this using integration by parts after making the substitution u=x+1u = x + 1, which expression correctly represents the transformed integral?

  1. ∫u2(u−1)1/2du\int u^2 (u-1)^{1/2} du
  2. ∫(u−1)2(u+1)1/2du\int (u-1)^2 (u+1)^{1/2} du
  3. ∫(u−1)2u1/2du\int (u-1)^2 u^{1/2} du (correct answer)
  4. ∫(u+1)2u1/2du\int (u+1)^2 u^{1/2} du
Explanation: When you encounter an integral with both polynomial and radical terms, substitution can simplify the expression before applying integration techniques. The key is carefully tracking how each part of the original integral transforms under the substitution. With the substitution u=x+1u = x + 1, we have x=u−1x = u - 1 and dx=dudx = du. Now we systematically replace each component of the original integral ∫x2x+1dx\int x^2 \sqrt{x+1} dx:
  • x2x^2 becomes (u−1)2(u-1)^2
  • x+1\sqrt{x+1} becomes u=u1/2\sqrt{u} = u^{1/2}
  • dxdx becomes dudu
This gives us ∫(u−1)2u1/2du\int (u-1)^2 u^{1/2} du, which matches answer C. Looking at the incorrect options: Answer A has u2u^2 instead of (u−1)2(u-1)^2, which would mean you forgot to substitute for x2x^2 and incorrectly wrote u2u^2. Answer B incorrectly transforms x+1\sqrt{x+1} to (u+1)1/2(u+1)^{1/2} when it should be u1/2u^{1/2}, and also has the wrong polynomial term. Answer D makes both errors from B: keeping u+1u+1 in the polynomial and radical parts, suggesting confusion about which direction the substitution goes. Study tip: When making substitutions, write out the relationships clearly (u=x+1u = x + 1 means x=u−1x = u - 1) and substitute each piece methodically. Double-check by verifying that your new variable appears consistently throughout the transformed integral.

Question 9

Consider the integral ∫(ln⁡x)2dx\int (\ln x)^2 dx. Using integration by parts with u=(ln⁡x)2u = (\ln x)^2, the resulting integral after one application is x(ln⁡x)2−2∫ln⁡xdxx(\ln x)^2 - 2\int \ln x dx. If ∫ln⁡xdx=xln⁡x−x+C\int \ln x dx = x \ln x - x + C, what is ∫(ln⁡x)2dx\int (\ln x)^2 dx?

  1. x(ln⁡x)2−2xln⁡x−2x+Cx(\ln x)^2 - 2x \ln x - 2x + C
  2. x(ln⁡x)2−2xln⁡x+2x+Cx(\ln x)^2 - 2x \ln x + 2x + C (correct answer)
  3. x(ln⁡x)2+2xln⁡x−2x+Cx(\ln x)^2 + 2x \ln x - 2x + C
  4. x(ln⁡x)2−xln⁡x+x+Cx(\ln x)^2 - x \ln x + x + C
Explanation: When you encounter integrals involving powers of logarithms, integration by parts is typically your go-to method. The key insight is recognizing that you'll often need to apply the technique multiple times or use previously known results. Starting with the given result after one integration by parts: x(ln⁡x)2−2∫ln⁡xdxx(\ln x)^2 - 2\int \ln x dx. Since you're provided that ∫ln⁡xdx=xln⁡x−x+C\int \ln x dx = x \ln x - x + C, you can substitute this directly into your expression. This gives you: x(ln⁡x)2−2(xln⁡x−x+C)=x(ln⁡x)2−2xln⁡x+2x−2Cx(\ln x)^2 - 2(x \ln x - x + C) = x(\ln x)^2 - 2x \ln x + 2x - 2C Since −2C-2C is just another constant, you can write the final answer as x(ln⁡x)2−2xln⁡x+2x+Cx(\ln x)^2 - 2x \ln x + 2x + C, which is choice B. Let's examine why the other options are incorrect: Choice A has −2x-2x instead of +2x+2x, which results from incorrectly distributing the negative sign when substituting ∫ln⁡xdx\int \ln x dx. Choice C has +2xln⁡x+2x \ln x instead of −2xln⁡x-2x \ln x, missing the negative sign from the −2-2 coefficient. Choice D appears to use an incorrect formula for ∫ln⁡xdx\int \ln x dx or makes errors in the algebraic manipulation. Remember this pattern: when using integration by parts with logarithmic functions, pay careful attention to sign changes when distributing coefficients. Always double-check your algebra when substituting known integral results, as sign errors are the most common mistakes in these problems.

Question 10

When evaluating ∫exsin⁡xdx\int e^x \sin x dx using integration by parts, after applying the technique twice, you obtain the equation ∫exsin⁡xdx=exsin⁡x−excos⁡x−∫exsin⁡xdx\int e^x \sin x dx = e^x \sin x - e^x \cos x - \int e^x \sin x dx. What is the correct next step?

  1. Add ∫exsin⁡xdx\int e^x \sin x dx to both sides and divide by 2 to solve for the integral (correct answer)
  2. Apply integration by parts a third time to the integral ∫exsin⁡xdx\int e^x \sin x dx on the right side
  3. Substitute u=sin⁡xu = \sin x and dv=exdxdv = e^x dx to restart the process with different choices
  4. Recognize that the equation is incorrect and restart with u=exu = e^x and dv=sin⁡xdxdv = \sin x dx
Explanation: This is the classic 'cyclic' integration by parts situation. When the same integral appears on both sides of the equation with opposite signs, we can solve algebraically. Adding ∫eˣsin x dx to both sides gives: 2∫eˣsin x dx = eˣsin x - eˣcos x, so ∫eˣsin x dx = (1/2)(eˣsin x - eˣcos x) + C. Choice B would create an infinite loop. Choice C would just restart the same process. Choice D is wrong because the equation shown is actually correct - this cyclic behavior is expected and useful for this type of integral.

Question 11

A student attempts to evaluate ∫0π/2xcos⁡xdx\int_0^{\pi/2} x \cos x dx using integration by parts with u=xu = x and dv=cos⁡xdxdv = \cos x dx. After applying the formula ∫udv=uv−∫vdu\int u dv = uv - \int v du, they get [xsin⁡x]0π/2−∫0π/2sin⁡xdx\left[ x \sin x \right]_0^{\pi/2} - \int_0^{\pi/2} \sin x dx. What is the value of this definite integral?

  1. π2+1\frac{\pi}{2} + 1
  2. π2−1\frac{\pi}{2} - 1 (correct answer)
  3. π2\frac{\pi}{2}
  4. 11
Explanation: When you encounter an integral involving a polynomial multiplied by a trigonometric function, integration by parts is typically your best strategy. The student has correctly set up the problem with u=xu = x and dv=cos⁡xdxdv = \cos x dx, giving du=dxdu = dx and v=sin⁡xv = \sin x. Let's complete the calculation. The student correctly applied the integration by parts formula to get: [xsin⁡x]0π/2−∫0π/2sin⁡xdx\left[ x \sin x \right]_0^{\pi/2} - \int_0^{\pi/2} \sin x dx First, evaluate the boundary term: [xsin⁡x]0π/2=π2sin⁡(π2)−0sin⁡(0)=π2⋅1−0=π2\left[ x \sin x \right]_0^{\pi/2} = \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) - 0 \sin(0) = \frac{\pi}{2} \cdot 1 - 0 = \frac{\pi}{2} Next, evaluate the remaining integral: ∫0π/2sin⁡xdx=[−cos⁡x]0π/2=−cos⁡(π2)−(−cos⁡(0))=0+1=1\int_0^{\pi/2} \sin x dx = \left[ -\cos x \right]_0^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 + 1 = 1 Therefore: π2−1\frac{\pi}{2} - 1 Choice A gives π2+1\frac{\pi}{2} + 1, which occurs if you incorrectly add instead of subtract the second integral. Choice C gives π2\frac{\pi}{2}, which happens if you forget to evaluate the second integral entirely. Choice D gives 11, which results from evaluating only the second integral while ignoring the boundary term. The correct answer is B. Study tip: In integration by parts problems, always double-check your signs and make sure you evaluate both the boundary term and the remaining integral completely. The LIPET rule (Logarithmic, Inverse trig, Polynomial, Exponential, Trigonometric) helps you choose uu wisely.

Question 12

If F(x)=∫1xt2ln⁡tdtF(x) = \int_1^x t^2 \ln t dt, what is F′(x)F'(x)?

  1. x2ln⁡x−x22x^2 \ln x - \frac{x^2}{2}
  2. x33ln⁡x−x39\frac{x^3}{3} \ln x - \frac{x^3}{9}
  3. x33ln⁡x−x39+19\frac{x^3}{3} \ln x - \frac{x^3}{9} + \frac{1}{9}
  4. x2ln⁡xx^2 \ln x (correct answer)
Explanation: When you see a function defined as a definite integral with a variable upper limit, you're dealing with the Fundamental Theorem of Calculus, Part I. This theorem tells us that if F(x)=∫axf(t)dtF(x) = \int_a^x f(t) dt, then F′(x)=f(x)F'(x) = f(x). To find F′(x)F'(x) for F(x)=∫1xt2ln⁡tdtF(x) = \int_1^x t^2 \ln t dt, you simply substitute xx for tt in the integrand. The integrand is t2ln⁡tt^2 \ln t, so F′(x)=x2ln⁡xF'(x) = x^2 \ln x. Choice D is correct: F′(x)=x2ln⁡xF'(x) = x^2 \ln x. Choice A gives x2ln⁡x−x22x^2 \ln x - \frac{x^2}{2}, which includes an unnecessary constant term. This suggests someone might have tried to evaluate the integral first, then differentiate—a much harder approach that's also incorrect here. Choices B and C, x33ln⁡x−x39\frac{x^3}{3} \ln x - \frac{x^3}{9} and x33ln⁡x−x39+19\frac{x^3}{3} \ln x - \frac{x^3}{9} + \frac{1}{9}, come from actually computing the antiderivative of t2ln⁡tt^2 \ln t using integration by parts, then differentiating. While these represent the antiderivative (with choice C including the constant from evaluating at the lower limit), the Fundamental Theorem gives us the derivative directly without needing to find the antiderivative first. Remember: When you see ddx∫axf(t)dt\frac{d}{dx}\int_a^x f(t) dt, don't integrate first—just replace tt with xx in the integrand. This saves time and avoids the complexity of integration by parts or other advanced techniques.

Question 13

Evaluate the definite integral ∫0π/2x2cos⁡(x) dx\int_{0}^{\pi/2} x^2 \cos(x) \,dx.

  1. π24−2\frac{\pi^2}{4} - 2 (correct answer)
  2. π24\frac{\pi^2}{4}
  3. π24+2\frac{\pi^2}{4} + 2
  4. π2−1\frac{\pi}{2} - 1
Explanation: This integral requires two applications of integration by parts. First, let u=x2u = x^2 and dv=cos⁡(x)dxdv = \cos(x) dx. Then du=2xdxdu = 2x dx and v=sin⁡(x)v = \sin(x). The integral becomes \[x^2 \sin(x)\]_{0}^{\pi/2} - \int_{0}^{\pi/2} 2x \sin(x) \,dx = (\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) - 0 - 2\int_{0}^{\pi/2} x \sin(x) \,dx = \frac{\pi^2}{4} - 2\int_{0}^{\pi/2} x \sin(x) \,dx. For the second integral, let u=xu = x and dv=sin⁡(x)dxdv = \sin(x) dx. Then du=dxdu = dx and v=−cos⁡(x)v = -\cos(x). The integral is \[-x \cos(x)\]_{0}^{\pi/2} - \int_{0}^{\pi/2} -\cos(x) \,dx = 0 + \int_{0}^{\pi/2} \cos(x) \,dx = [\sin(x)]_{0}^{\pi/2} = 1. Substituting back gives π24−2(1)=π24−2\frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2.

Question 14

If f(x)f(x) is an integrable function such that ∫f(x)ex dx=f(x)ex−∫3x2ex dx\int f(x) e^x \,dx = f(x)e^x - \int 3x^2 e^x \,dx, what is f(x)f(x)?

  1. x3x^3 (correct answer)
  2. 3x23x^2
  3. exe^x
  4. x3exx^3 e^x
Explanation: The formula for integration by parts is ∫u dv=uv−∫v du\int u \,dv = uv - \int v \,du. The given equation is ∫f(x)ex dx=f(x)ex−∫3x2ex dx\int f(x) e^x \,dx = f(x)e^x - \int 3x^2 e^x \,dx. We can match this to the IBP formula by identifying the parts. The original integral is ∫u dv\int u \,dv. Let's set u=f(x)u=f(x) and dv=exdxdv=e^x dx. Then v=∫ex dx=exv = \int e^x \,dx = e^x. The right side of the IBP formula is uv−∫v du=f(x)ex−∫exf′(x) dxuv - \int v \,du = f(x)e^x - \int e^x f'(x) \,dx. Comparing this to the given expression, we see that ∫exf′(x) dx=∫3x2ex dx\int e^x f'(x) \,dx = \int 3x^2 e^x \,dx. This implies that f′(x)=3x2f'(x) = 3x^2. Integrating f′(x)f'(x) to find f(x)f(x) gives f(x)=∫3x2 dx=x3f(x) = \int 3x^2 \,dx = x^3 (we can ignore the constant of integration here).

Question 15

Evaluate ∫xsin⁡(x)cos⁡(x) dx\int x \sin(x) \cos(x) \,dx.

  1. −x4cos⁡(2x)+18sin⁡(2x)+C-\frac{x}{4}\cos(2x) + \frac{1}{8}\sin(2x) + C (correct answer)
  2. x24sin⁡2(x)+C\frac{x^2}{4}\sin^2(x) + C
  3. −x2cos⁡(2x)+14sin⁡(2x)+C-\frac{x}{2}\cos(2x) + \frac{1}{4}\sin(2x) + C
  4. x2sin⁡(2x)−14cos⁡(2x)+C\frac{x}{2}\sin(2x) - \frac{1}{4}\cos(2x) + C
Explanation: First, simplify the integrand using the double angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x), so sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x). The integral becomes 12∫xsin⁡(2x) dx\frac{1}{2}\int x \sin(2x) \,dx. Now use integration by parts with u=xu=x and dv=sin⁡(2x)dxdv=\sin(2x)dx. Then du=dxdu=dx and v=−12cos⁡(2x)v = -\frac{1}{2}\cos(2x). The integral is 12(x(−12cos⁡(2x))−∫−12cos⁡(2x) dx)=12(−x2cos⁡(2x)+12∫cos⁡(2x) dx)\frac{1}{2} \left( x(-\frac{1}{2}\cos(2x)) - \int -\frac{1}{2}\cos(2x) \,dx \right) = \frac{1}{2} \left( -\frac{x}{2}\cos(2x) + \frac{1}{2} \int \cos(2x) \,dx \right). Integrating the final term gives 12(−x2cos⁡(2x)+12⋅12sin⁡(2x))+C=−x4cos⁡(2x)+18sin⁡(2x)+C\frac{1}{2} \left( -\frac{x}{2}\cos(2x) + \frac{1}{2} \cdot \frac{1}{2}\sin(2x) \right) + C = -\frac{x}{4}\cos(2x) + \frac{1}{8}\sin(2x) + C.

Question 16

The application of integration by parts to an integral ∫f(x) dx\int f(x) \,dx yields the expression x2sin⁡(x)−∫2xsin⁡(x) dxx^2 \sin(x) - \int 2x \sin(x) \,dx. What was the original integral ∫f(x) dx\int f(x) \,dx?

  1. ∫x2cos⁡(x) dx\int x^2 \cos(x) \,dx (correct answer)
  2. ∫x2sin⁡(x) dx\int x^2 \sin(x) \,dx
  3. ∫2xcos⁡(x) dx\int 2x \cos(x) \,dx
  4. ∫2xsin⁡(x) dx\int 2x \sin(x) \,dx
Explanation: The integration by parts formula is ∫u dv=uv−∫v du\int u \,dv = uv - \int v \,du. We are given uv−∫v du=x2sin⁡(x)−∫2xsin⁡(x) dxuv - \int v \,du = x^2 \sin(x) - \int 2x \sin(x) \,dx. By comparing the terms, we can identify uv=x2sin⁡(x)uv = x^2 \sin(x) and ∫v du=∫2xsin⁡(x) dx\int v \,du = \int 2x \sin(x) \,dx. From v du=2xsin⁡(x) dxv \,du = 2x \sin(x) \,dx, let's test the components from uvuv. If we let u=x2u=x^2, then v=sin⁡(x)v=\sin(x). This means du=2x dxdu = 2x \,dx and dv=cos⁡(x) dxdv=\cos(x) \,dx. Let's check if these match v duv \,du: v du=(sin⁡x)(2x dx)v \,du = (\sin x)(2x \,dx), which matches. The original integral is ∫u dv\int u \,dv, which is ∫x2cos⁡(x) dx\int x^2 \cos(x) \,dx.

Question 17

Evaluate the improper integral ∫0∞xe−2x dx\int_{0}^{\infty} x e^{-2x} \,dx.

  1. 1/41/4 (correct answer)
  2. 1/21/2
  3. −1/4-1/4
  4. The integral diverges.
Explanation: First, find the indefinite integral ∫xe−2x dx\int x e^{-2x} \,dx using integration by parts. Let u=xu=x and dv=e−2xdxdv=e^{-2x}dx. Then du=dxdu=dx and v=−12e−2xv = -\frac{1}{2}e^{-2x}. The antiderivative is −x2e−2x−∫−12e−2x dx=−x2e−2x−14e−2x-\frac{x}{2}e^{-2x} - \int -\frac{1}{2}e^{-2x} \,dx = -\frac{x}{2}e^{-2x} - \frac{1}{4}e^{-2x}. Now, evaluate the improper integral: lim⁡b→∞[−x2e−2x−14e−2x]0b\lim_{b \to \infty} [-\frac{x}{2}e^{-2x} - \frac{1}{4}e^{-2x}]_{0}^{b}. This is lim⁡b→∞((−b2e−2b−14e−2b)−(0−14e0))\lim_{b \to \infty} ((-\frac{b}{2}e^{-2b} - \frac{1}{4}e^{-2b}) - (0 - \frac{1}{4}e^{0})). As b→∞b \to \infty, both e−2be^{-2b} and be−2bbe^{-2b} (by L'Hôpital's Rule) approach 0. So the expression becomes 0−(−14))=140 - (-\frac{1}{4})) = \frac{1}{4}.

Question 18

Evaluate ∫xcosh⁡(x) dx\int x \cosh(x) \,dx.

  1. xsinh⁡(x)−cosh⁡(x)+Cx \sinh(x) - \cosh(x) + C (correct answer)
  2. xsinh⁡(x)+cosh⁡(x)+Cx \sinh(x) + \cosh(x) + C
  3. −xsinh⁡(x)+cosh⁡(x)+C-x \sinh(x) + \cosh(x) + C
  4. x22sinh⁡(x)+C\frac{x^2}{2}\sinh(x) + C
Explanation: This integral is analogous to ∫xcos⁡(x)dx\int x \cos(x) dx. We use integration by parts with u=xu=x and dv=cosh⁡(x)dxdv = \cosh(x) dx. Then du=dxdu=dx and v=sinh⁡(x)v = \sinh(x) (since the derivative of sinh⁡(x)\sinh(x) is cosh⁡(x)\cosh(x)). The formula gives uv−∫v du=xsinh⁡(x)−∫sinh⁡(x) dxuv - \int v \,du = x \sinh(x) - \int \sinh(x) \,dx. Since the integral of sinh⁡(x)\sinh(x) is cosh⁡(x)\cosh(x), the final result is xsinh⁡(x)−cosh⁡(x)+Cx \sinh(x) - \cosh(x) + C. Distractors B and C contain sign errors based on confusion with trigonometric function integrals.

Question 19

Evaluate the definite integral ∫01x(1−x)5 dx\int_{0}^{1} x(1-x)^5 \,dx.

  1. 142\frac{1}{42} (correct answer)
  2. −142-\frac{1}{42}
  3. 130\frac{1}{30}
  4. 00
Explanation: This can be solved efficiently with integration by parts. Let u=xu=x and dv=(1−x)5dxdv=(1-x)^5 dx. Then du=dxdu=dx and v=−(1−x)66v = -\frac{(1-x)^6}{6}. The integral is \[-\frac{x(1-x)^6}{6}\]_{0}^{1} - \int_{0}^{1} -\frac{(1-x)^6}{6} \,dx. The first term evaluates to (−1(0)66)−(−0(1)66)=0(-\frac{1(0)^6}{6}) - (-\frac{0(1)^6}{6}) = 0. The integral part becomes 16∫01(1−x)6 dx=16[−(1−x)77]01=16((0)−(−(1)77))=16(17)=142\frac{1}{6}\int_{0}^{1}(1-x)^6 \,dx = \frac{1}{6} [-\frac{(1-x)^7}{7}]_{0}^{1} = \frac{1}{6} ( (0) - (-\frac{(1)^7}{7}) ) = \frac{1}{6} (\frac{1}{7}) = \frac{1}{42}. Alternatively, a substitution w=1−xw=1-x also works well.

Question 20

Evaluate ∫arctan⁡(1/x) dx\int \arctan(1/x) \,dx.

  1. xarctan⁡(1/x)−12ln⁡(x2+1)+Cx \arctan(1/x) - \frac{1}{2}\ln(x^2+1) + C
  2. xarctan⁡(1/x)+ln⁡(x2+1)+Cx \arctan(1/x) + \ln(x^2+1) + C
  3. xarctan⁡(1/x)+12ln⁡(x2+1)+Cx \arctan(1/x) + \frac{1}{2}\ln(x^2+1) + C (correct answer)
  4. arctan⁡(1/x)−xx2+1+C\arctan(1/x) - \frac{x}{x^2+1} + C
Explanation: Use integration by parts with u=arctan⁡(1/x)u = \arctan(1/x) and dv=dxdv = dx. For the derivative of uu, we use the chain rule: du=11+(1/x)2⋅(−1x2) dx=1(x2+1)/x2⋅(−1x2) dx=−1x2+1 dxdu = \frac{1}{1+(1/x)^2} \cdot (-\frac{1}{x^2}) \,dx = \frac{1}{(x^2+1)/x^2} \cdot (-\frac{1}{x^2}) \,dx = -\frac{1}{x^2+1} \,dx. We have v=xv=x. Applying the formula: ∫arctan⁡(1/x) dx=xarctan⁡(1/x)−∫x(−1x2+1) dx=xarctan⁡(1/x)+∫xx2+1 dx\int \arctan(1/x) \,dx = x\arctan(1/x) - \int x(-\frac{1}{x^2+1}) \,dx = x\arctan(1/x) + \int \frac{x}{x^2+1} \,dx. The remaining integral can be solved with a substitution w=x2+1w=x^2+1, dw=2xdxdw=2x dx, which gives 12ln⁡(x2+1)\frac{1}{2}\ln(x^2+1). The final answer is xarctan⁡(1/x)+12ln⁡(x2+1)+Cx \arctan(1/x) + \frac{1}{2}\ln(x^2+1) + C.