AP Calculus BC Practice Quiz: Integrating Using Integration by Parts (BC ONLY)

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Evaluate the indefinite integral ∫x sin(x) dx.

A)-x cos(x) + sin(x) + CB)x cos(x) - sin(x) + CC)-(x^2/2) cos(x) + CD)x sin(x) - cos(x) + C

All Questions (7)

Evaluate the indefinite integral ∫x sin(x) dx.

A) -x cos(x) + sin(x) + C

B) x cos(x) - sin(x) + C

C) -(x^2/2) cos(x) + C

D) x sin(x) - cos(x) + C

Correct Answer: A

To solve ∫x sin(x) dx using integration by parts (∫u dv = uv - ∫v du), let u = x and dv = sin(x) dx. Then, du = dx and v = -cos(x). Plugging these into the formula gives: x(-cos(x)) - ∫(-cos(x)) dx = -x cos(x) + ∫cos(x) dx = -x cos(x) + sin(x) + C.

Evaluate the definite integral ∫ from 0 to 1 of x * e^(2x) dx.

A) (e^2 + 1) / 4

B) (e^2 - 1) / 2

C) e^2 / 2

D) (e^2 + 1) / 2

Correct Answer: A

Using integration by parts, let u = x and dv = e^(2x) dx. Then du = dx and v = (1/2)e^(2x). The antiderivative is uv - ∫v du = x(1/2)e^(2x) - ∫(1/2)e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x). Evaluating this from 0 to 1: [(1/2)(1)e^2 - (1/4)e^2] - [(1/2)(0)e^0 - (1/4)e^0] = [(1/2)e^2 - (1/4)e^2] - [0 - 1/4] = (1/4)e^2 + 1/4 = (e^2 + 1) / 4.

Which of the following is the result of evaluating ∫ln(x) dx?

A) 1/x + C

B) x ln(x) - x + C

C) x ln(x) + x + C

D) ln(x) - 1 + C

Correct Answer: B

This integral requires integration by parts. Let u = ln(x) and dv = dx. Then du = (1/x) dx and v = x. Using the formula ∫u dv = uv - ∫v du, we get: ln(x) * x - ∫x * (1/x) dx = x ln(x) - ∫1 dx = x ln(x) - x + C.

To evaluate the integral ∫x^2 cos(x) dx using integration by parts, the technique must be applied twice. What is the resulting antiderivative?

A) x^2 sin(x) - 2x cos(x) - 2sin(x) + C

B) x^2 sin(x) + 2x cos(x) - 2sin(x) + C

C) -x^2 sin(x) + 2x cos(x) + 2sin(x) + C

D) 2x sin(x) + x^2 cos(x) + C

Correct Answer: B

First application: let u = x^2, dv = cos(x) dx. Then du = 2x dx, v = sin(x). This gives x^2 sin(x) - ∫2x sin(x) dx. Second application for ∫2x sin(x) dx: let u = 2x, dv = sin(x) dx. Then du = 2 dx, v = -cos(x). This gives 2x(-cos(x)) - ∫-cos(x)(2) dx = -2x cos(x) + 2∫cos(x) dx = -2x cos(x) + 2sin(x). Substituting back: x^2 sin(x) - (-2x cos(x) + 2sin(x)) + C = x^2 sin(x) + 2x cos(x) - 2sin(x) + C.

When evaluating ∫x arctan(x) dx using integration by parts (∫u dv = uv - ∫v du), which of the following is the most effective choice for u and dv?

A) u = x, dv = arctan(x) dx

B) u = 1, dv = x arctan(x) dx

C) u = x arctan(x), dv = dx

D) u = arctan(x), dv = x dx

Correct Answer: D

The most effective strategy for choosing 'u' often follows the LIATE/LIPET rule (Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential). Since the integrand contains an Inverse Trigonometric function (arctan(x)) and an Algebraic function (x), we should choose u to be the inverse trigonometric function. Therefore, u = arctan(x) and dv = x dx. This is because the derivative of arctan(x) is a simple algebraic function (1/(1+x^2)), while its integral is very complex.

Evaluate ∫ from 1 to e of x^2 ln(x) dx.

A) (e^3 + 1) / 9

B) (2e^3 - 1) / 9

C) (2e^3 + 1) / 9

D) e^3 / 3

Correct Answer: C

Use integration by parts with u = ln(x) and dv = x^2 dx. Then du = (1/x) dx and v = x^3/3. The antiderivative is uv - ∫v du = (x^3/3)ln(x) - ∫(x^3/3)(1/x) dx = (x^3/3)ln(x) - ∫(x^2/3) dx = (x^3/3)ln(x) - x^3/9. Now, evaluate from 1 to e: [(e^3/3)ln(e) - e^3/9] - [(1^3/3)ln(1) - 1^3/9] = [(e^3/3)(1) - e^3/9] - [(1/3)(0) - 1/9] = [3e^3/9 - e^3/9] - [-1/9] = 2e^3/9 + 1/9 = (2e^3 + 1) / 9.

The integral ∫e^x cos(x) dx is a classic example of an integral that requires applying integration by parts twice and then solving algebraically for the integral. What is the result?

A) (e^x / 2)(sin(x) + cos(x)) + C

B) e^x (sin(x) - cos(x)) + C

C) (e^x / 2)(sin(x) - cos(x)) + C

D) e^x sin(x) + C

Correct Answer: A

Let I = ∫e^x cos(x) dx. First, let u = e^x, dv = cos(x) dx. Then du = e^x dx, v = sin(x). So, I = e^x sin(x) - ∫e^x sin(x) dx. For the second integral, let u = e^x, dv = sin(x) dx. Then du = e^x dx, v = -cos(x). So, ∫e^x sin(x) dx = -e^x cos(x) - ∫(-cos(x))e^x dx = -e^x cos(x) + I. Substitute back: I = e^x sin(x) - [-e^x cos(x) + I]. This simplifies to I = e^x sin(x) + e^x cos(x) - I. Solving for I: 2I = e^x(sin(x) + cos(x)), so I = (e^x / 2)(sin(x) + cos(x)) + C.