AP Calculus BC Quiz: Integrating Using Integration By Parts

What this quiz covers

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

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.

Question 1

1. $e^{x}(x+1)+C$
2. $e^{x}(x-1)+C$ (correct answer)
3. $xe^{x}-\int e^{x}\,dx$
4. $\frac{x^{2}}{2}e^{x}+C$
5. $e^{x}+\frac{x^{2}}{2}+C$

Explanation: This integral requires integration by parts to handle the product of x and e^x. For integration by parts, we use ∫u dv = uv - ∫v du, choosing u = x (so du = dx) and dv = e^x dx (so v = e^x). Applying the formula: ∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x + C = e^x(x-1) + C. Choice C shows the incomplete form before finishing the integration, which is a common error when students forget to complete the final integral. When using integration by parts with polynomial times exponential, always let u be the polynomial part to reduce its degree.

Question 2

1. $\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$ (correct answer)
2. $\frac{2}{3}x^{3/2}\ln x+\frac{4}{9}x^{3/2}+C$
3. $\frac{2}{3}x^{3/2}(\ln x)^2+C$
4. $x^{1/2}\ln x-2x^{1/2}+C$
5. $\frac{3}{2}x^{1/2}\ln x-\frac{3}{2}x^{1/2}+C$

Explanation: This problem requires integration by parts to find the antiderivative of \sqrt{x} ln x for x > 0. To apply integration by parts, choose u = ln x because it differentiates to 1/x, and dv = x^{1/2} dx because it integrates to (2/3) x^{3/2}. Then du = (1/x) dx and v = (2/3) x^{3/2}. Substituting into the formula gives (2/3) x^{3/2} ln x - ∫ (2/3) x^{3/2} (1/x) dx, which simplifies to (2/3) x^{3/2} ln x - (4/9) x^{3/2} + C. A tempting distractor might be (2/3) x^{3/2} ln x + (4/9) x^{3/2} + C, from neglecting the negative sign in parts. A transferable strategy for integration by parts is to select u as the log for rational derivative, and dv as the power function, adjusting exponents carefully in the process.

Question 3

1. $-e^{-x}(x+1)+C$ (correct answer)
2. $-e^{-x}(x-1)+C$
3. $e^{-x}(x+1)+C$
4. $xe^{-x}-\int e^{-x}\,dx$
5. $\frac{x^{2}}{2}e^{-x}+C$

Explanation: This integral requires integration by parts for xe^(-x). Choose u = x (so du = dx) and dv = e^(-x) dx (so v = -e^(-x)), noting the negative from integrating e^(-x). Applying the formula: ∫xe^(-x) dx = x(-e^(-x)) - ∫(-e^(-x)) dx = -xe^(-x) + ∫e^(-x) dx = -xe^(-x) - e^(-x) + C = -e^(-x)(x + 1) + C. Choice C has the wrong overall sign, missing the negative that comes from integrating e^(-x). When integrating products with e^(-x), remember that ∫e^(-x) dx = -e^(-x), introducing a crucial negative sign.

Question 4

1. $\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}+C$ (correct answer)
2. $\tfrac{x^2}{2}\ln x+\tfrac{x^2}{4}+C$
3. $x\ln x-x+C$
4. $\tfrac{x^2}{2}\ln x-\tfrac{x}{2}+C$
5. $\tfrac{(\ln x)^2}{2}+C$

Explanation: Integration by parts is the key skill for evaluating ∫ x ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = x dx so v = (1/2) x^2. This choice works well as ln x simplifies upon differentiation, pairing with the polynomial growth in dv. The result is (1/2) x^2 ln x - ∫ (1/2) x^2 (1/x) dx, simplifying to (1/2) x^2 ln x - (1/4) x^2 + C. A tempting distractor like choice B fails with a positive (1/4) x^2 term, likely from neglecting the subtraction in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 5

1. $-\tfrac{\ln x+1}{x}+C$ (correct answer)
2. $-\tfrac{\ln x-1}{x}+C$
3. $\tfrac{\ln x+1}{x}+C$
4. $-\tfrac{(\ln x)^2}{2}+C$
5. $\tfrac{1}{x}+C$

Explanation: Integration by parts is the key skill for evaluating ∫ (ln x)/x^2 dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = x^{-2} dx so v = -x^{-1}. This selection simplifies the logarithmic part, leaving a rational integral. The computation yields - (ln x)/x - ∫ - (1/x) (1/x) dx, simplifying to - (ln x + 1)/x + C. A tempting distractor like choice C fails by having a positive overall sign, possibly from forgetting the negative v in the initial product. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 6

1. $x\ln x - x + C$ (correct answer)
2. $x\ln x + x + C$
3. $\tfrac{(\ln x)^2}{2}+C$
4. $\ln x - x + C$
5. $\tfrac{1}{x}+C$

Explanation: Integration by parts is the key skill for evaluating ∫ ln x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = ln x so du = (1/x) dx, and dv = dx so v = x. This selection is ideal because ln x differentiates to a simple rational function, while dv is basic. The result is x ln x - ∫ x (1/x) dx, simplifying to x ln x - x + C. A tempting distractor like choice B fails with a positive x term, which would result from forgetting the negative sign in the parts formula. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 7

1. $e^{x}(x^{2}-2x+2)+C$ (correct answer)
2. $e^{x}(x^{2}+2x+2)+C$
3. $\frac{x^{3}}{3}e^{x}+C$
4. $x^{2}e^{x}-\int 2xe^{x}\,dx$
5. $e^{x}x^{2}-2e^{x}+C$

Explanation: This integral requires integration by parts twice for $x^2 e^x$. First, let $u = x^2$ (so $du = 2x \, dx$) and $dv = e^x \, dx$ (so $v = e^x$), giving $∫ x^2 e^x \, dx = x^2 e^x - ∫ 2x e^x \, dx$. For the remaining integral $∫ 2x e^x \, dx$, apply parts again with $u = 2x$ and $dv = e^x \, dx$, yielding $2x e^x - 2 e^x$. Combining: $∫ x^2 e^x \, dx = x^2 e^x - 2x e^x + 2 e^x + C = e^x (x^2 - 2x + 2) + C$. Choice D shows only the first application of parts without completing the process. For polynomial times exponential, repeated integration by parts reduces the polynomial degree to zero.

Question 8

1. $\frac{e^{x}}{2}(\sin x-\cos x)+C$
2. $\frac{e^{x}}{2}(\sin x+\cos x)+C$ (correct answer)
3. $e^{x}\sin x+C$
4. $e^{x}\cos x-\int e^{x}(-\sin x)\,dx$
5. $\sin x+e^{x}+C$

Explanation: This integral ∫e^x cos x dx requires integration by parts twice, creating a cyclic pattern. First, let u = cos x and dv = e^x dx, giving ∫e^x cos x dx = e^x cos x - ∫e^x(-sin x) dx = e^x cos x + ∫e^x sin x dx. For the new integral, use u = sin x and dv = e^x dx: ∫e^x sin x dx = e^x sin x - ∫e^x cos x dx. This gives us back our original integral, so combining both equations and solving: 2∫e^x cos x dx = e^x cos x + e^x sin x, thus ∫e^x cos x dx = (e^x/2)(sin x + cos x) + C. Choice D shows the incomplete first step without recognizing the cyclic pattern. For exponential times trig functions, expect to use parts twice and solve algebraically.

Question 9

1. $-x^2\cos x+2x\sin x+2\cos x+C$ (correct answer)
2. $x^2\cos x-2x\sin x-2\cos x+C$
3. $-x^2\cos x+2x\sin x-2\cos x+C$
4. $-x^2\cos x+\sin x+C$
5. $2x\sin x+2\cos x+C$

Explanation: Integration by parts is the key skill for evaluating ∫ x^2 sin x dx, based on the formula ∫ u dv = u v - ∫ v du, often applied repeatedly. Choose u = x^2 so du = 2x dx, and dv = sin x dx so v = -cos x for the first step. This leads to -x^2 cos x + 2 ∫ x cos x dx, requiring another parts application where u = x and dv = cos x dx. The full computation yields -x^2 cos x + 2x sin x + 2 cos x + C after integrating twice. A tempting distractor like choice C fails by having a negative 2 cos x term, possibly from a sign error in the second integration step. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 10

1. $\frac{x e^{2x}}{2}+\frac{e^{2x}}{4}+C$
2. $\frac{x e^{2x}}{2}-\frac{e^{2x}}{4}+C$ (correct answer)
3. $\frac{x e^{2x}}{2}-\frac{e^{2x}}{2}+C$
4. $x e^{2x}-\frac{e^{2x}}{2}+C$
5. $\frac{x^2 e^{2x}}{2}+C$

Explanation: This problem requires integration by parts to find the antiderivative of x e^{2x}. To apply integration by parts, choose u = x because its derivative simplifies to 1, and dv = e^{2x} dx because exponentials are straightforward to integrate. Then du = dx and v = (1/2) e^{2x}. Substituting into the formula gives (x/2) e^{2x} - ∫ (1/2) e^{2x} dx, which evaluates to (x/2) e^{2x} - (1/4) e^{2x} + C. A tempting distractor might be the expression with a plus sign instead of minus, resulting from omitting the negative sign in the parts formula. A transferable strategy for integration by parts is to let u be the polynomial term that decreases in degree when differentiated, and dv the exponential or transcendental function that integrates easily.

Question 11

1. $\tfrac{1}{2}xe^{2x}-\tfrac{1}{4}e^{2x}+C$ (correct answer)
2. $\tfrac{1}{2}xe^{2x}+\tfrac{1}{4}e^{2x}+C$
3. $\tfrac{1}{2}x^2e^{2x}+C$
4. $\tfrac{1}{4}x e^{2x}-\tfrac{1}{8}e^{2x}+C$
5. $e^{2x}(x-1)+C$

Explanation: Integration by parts is the key skill for evaluating ∫ x e^{2x} dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = x so du = dx, and dv = e^{2x} dx so v = (1/2) e^{2x}. This selection works because differentiating x simplifies the integrand, while integrating the exponential is straightforward. Applying the formula yields (1/2) x e^{2x} - ∫ (1/2) e^{2x} dx, which simplifies to (1/2) x e^{2x} - (1/4) e^{2x} + C. A tempting distractor like choice B fails because it has a positive (1/4) e^{2x} term, which would arise from an incorrect sign in the subtraction step. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 12

1. $\tfrac{1}{2}e^x(\sin x-\cos x)+C$ (correct answer)
2. $\tfrac{1}{2}e^x(\sin x+\cos x)+C$
3. $e^x\sin x+C$
4. $\tfrac{1}{2}e^x(\cos x-\sin x)+C$
5. $-\tfrac{1}{2}e^x(\sin x+\cos x)+C$

Explanation: Integration by parts is the key skill for evaluating ∫ e^x sin x dx, based on the formula ∫ u dv = u v - ∫ v du, applied twice to cycle back. Choose u = sin x so du = cos x dx, and dv = e^x dx so v = e^x for the first step. This produces e^x sin x - ∫ e^x cos x dx, then repeat with u = cos x and dv = e^x dx. Solving the system yields (1/2) e^x (sin x - cos x) + C. A tempting distractor like choice B fails by having a positive cos x term, which might come from a sign error in resolving the cyclic integrals. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 13

1. $x\cos x-\sin x+C$
2. $-x\cos x+\sin x+C$ (correct answer)
3. $x\sin x+\cos x+C$
4. $-x\sin x-\cos x+C$
5. $\sin x+C$

Explanation: For this integral of x sin(x), we apply integration by parts with u = x and dv = sin(x) dx. This gives du = dx and v = -cos(x) after integrating sin(x). Using the parts formula: ∫x sin(x) dx = x(-cos(x)) - ∫(-cos(x)) dx = -x cos(x) + ∫cos(x) dx = -x cos(x) + sin(x) + C. Option A has the wrong signs throughout, showing x cos(x) - sin(x) instead of -x cos(x) + sin(x). When integrating products involving sine or cosine, remember that ∫sin(x) dx = -cos(x) and ∫cos(x) dx = sin(x), and track signs carefully through each step.

Question 14

1. $x\tan x-\ln|\sec x|+C$ (correct answer)
2. $x\tan x+\ln|\sec x|+C$
3. $\tan x+\ln|\sec x|+C$
4. $x\sec^2 x-\tan x+C$
5. $\tfrac{x^2}{2}\sec^2 x+C$

Explanation: Integration by parts is the key skill for evaluating ∫ x sec^2 x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = x so du = dx, and dv = sec^2 x dx so v = tan x. This selection is effective because the polynomial x simplifies, leaving a trigonometric integral. The computation gives x tan x - ∫ tan x dx, which integrates to x tan x - ln|sec x| + C. A tempting distractor like choice B fails with a positive ln|sec x| term, possibly from inverting the sign when integrating tan x. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 15

1. $x\arctan x-\tfrac{1}{2}\ln(1+x^2)+C$ (correct answer)
2. $x\arctan x+\tfrac{1}{2}\ln(1+x^2)+C$
3. $\tfrac{1}{1+x^2}+C$
4. $\arctan x-\tfrac{1}{2}\ln(1+x^2)+C$
5. $\tfrac{1}{2}(\arctan x)^2+C$

Explanation: Integration by parts is the key skill for evaluating ∫ arctan x dx, based on the formula ∫ u dv = u v - ∫ v du. Choose u = arctan x so du = 1/(1+x^2) dx, and dv = dx so v = x. This choice leverages the derivative of arctan x, which pairs nicely with the remaining integral. The result is x arctan x - ∫ x/(1+x^2) dx, simplifying to x arctan x - (1/2) ln(1+x^2) + C. A tempting distractor like choice B fails with a positive ln term, which could stem from a sign mistake in the parts subtraction. A transferable strategy for integration by parts is to prioritize u as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 16

1. $e^x(x^3-3x^2+6x-6)+C$ (correct answer)
2. $e^x(x^3+3x^2+6x+6)+C$
3. $e^x(x^3-3x^2-6x-6)+C$
4. $x^3e^x+C$
5. $\tfrac{1}{4}x^4e^x+C$

Explanation: Integration by parts is the key skill for evaluating $\int x^3 e^x \, dx$, based on the formula $\int u \, dv = u v - \int v \, du$, repeated for the polynomial. Choose $u = x^3$ so $du = 3x^2 \, dx$, and $dv = e^x \, dx$ so $v = e^x$, then repeat for lower powers. This tabular approach reduces the degree each time, leading to subsequent integrals of $x^2 e^x$, $x e^x$, and $e^x$. The full expansion yields $e^x (x^3 - 3x^2 + 6x - 6) + C$. A tempting distractor like choice B fails with all positive coefficients, likely from consistent sign errors in the repeated subtractions. A transferable strategy for integration by parts is to prioritize $u$ as the part that simplifies most upon differentiation, such as polynomials over exponentials or logarithms.

Question 17

1. $\frac{1}{2}e^x(\cos x-\sin x)+C$
2. $e^x(\cos x+\sin x)+C$
3. $\frac{1}{2}e^x(\cos x+\sin x)+C$ (correct answer)
4. $\frac{1}{2}e^x(\sin x-\cos x)+C$
5. $e^x\sin x+C$

Explanation: This problem requires integration by parts to find the antiderivative of e^x cos x, applied twice. To apply integration by parts, choose u = cos x and dv = e^x dx for the first step, as the exponential integrates to itself; then du = -sin x dx and v = e^x. This leads to e^x cos x + ∫ e^x sin x dx; repeating with u = sin x and dv = e^x dx gives e^x sin x - ∫ e^x cos x dx, resulting in the original integral on both sides and solving to (1/2) e^x (cos x + sin x) + C. A tempting distractor might be without the 1/2 factor, from forgetting to divide after solving the equation. A transferable strategy for integration by parts is to alternate choices of u and dv for products of exponentials and trig functions, leading to a solvable equation when the integral reappears.

Question 18

1. $\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$ (correct answer)
2. $\frac{x^3}{3}(\ln x)^2+C$
3. $\frac{x^3}{3}\ln x+\frac{x^3}{9}+C$
4. $\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$
5. $x^2\ln x-2x+C$

Explanation: This problem requires integration by parts to find the antiderivative of x^2 ln x for x > 0. To apply integration by parts, choose u = ln x because it differentiates to 1/x, and dv = x^2 dx because it integrates to (1/3) x^3. Then du = (1/x) dx and v = (1/3) x^3. Substituting into the formula gives (1/3) x^3 ln x - ∫ (1/3) x^3 (1/x) dx, which simplifies to (1/3) x^3 ln x - (1/9) x^3 + C. A tempting distractor might be (1/3) x^3 ln x + (1/9) x^3 + C, stemming from a sign error in the parts formula. A transferable strategy for integration by parts is to select u as the logarithm to simplify, and dv as the higher-power polynomial to integrate.

Question 19

1. $x\ln(2x)-x+C$ (correct answer)
2. $\ln(2x)+C$
3. $x\ln(2x)+x+C$
4. $\frac{(\ln(2x))^2}{2}+C$
5. $\frac{\ln(2x)}{x}+C$

Explanation: This problem requires integration by parts to find the antiderivative of ln(2x) for x > 0. To apply integration by parts, choose u = ln(2x) because its derivative is 1/x, and dv = dx because it integrates to x. Then du = (1/x) dx and v = x. Substituting into the formula gives x ln(2x) - ∫ x (1/x) dx, which simplifies to x ln(2x) - x + C. A tempting distractor might be x ln(2x) + x + C, which could arise from mishandling the sign in the subtraction step. A transferable strategy for integration by parts is to let u be the logarithmic term for differentiation to a rational function, and dv the constant or simple term for easy integration.

Question 20

1. $\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$ (correct answer)
2. $x\ln x-x+C$
3. $\frac{x^2}{2}\ln x-\frac{x}{2}+C$
4. $\frac{x^2}{2}(\ln x)^2+C$
5. $\frac{x^2}{2}\ln x+\frac{x^2}{4}+C$

Explanation: This problem requires integration by parts to find the antiderivative of x ln x for x > 0. To apply integration by parts, choose u = ln x because its derivative is 1/x, and dv = x dx because polynomials integrate easily. Then du = (1/x) dx and v = (1/2) x^2. Substituting into the formula gives (1/2) x^2 ln x - ∫ (1/2) x^2 (1/x) dx, which simplifies to (1/2) x^2 ln x - (1/4) x^2 + C. A tempting distractor might be an expression like (1/2) x^2 ln x + (1/4) x^2 + C, which arises from flipping the sign after integration. A transferable strategy for integration by parts is to let u be the logarithmic function since it simplifies upon differentiation, and dv the polynomial to integrate directly.