Integration by Parts Practice Problems with Answers

These integration by parts practice problems start with one round of parts and build up to repeated parts, the tabular method and integrals that loop back to themselves. Try each one first. The answers come first so you can check quickly, then the full solutions.

The formula comes from the product rule:

Pick u and dv, find du and v, then fill in the formula. For more on the formula, read the integration by parts rule.

Problems

One round of parts

1. \displaystyle\int x\,e^x\,dx
2. \displaystyle\int x\cos x\,dx
3. \displaystyle\int \ln x\,dx
4. \displaystyle\int x\ln x\,dx
5. \displaystyle\int x\sin(2x)\,dx
6. \displaystyle\int \arctan x\,dx

Repeated parts

7. \displaystyle\int x^2 e^x\,dx
8. \displaystyle\int x^2\cos x\,dx

Integrals that loop back

9. \displaystyle\int e^x\sin x\,dx
10. \displaystyle\int e^{2x}\cos x\,dx

Definite integrals

11. \displaystyle\int_0^1 x\,e^x\,dx
12. \displaystyle\int_1^{e} x\ln x\,dx

Answers

Step-by-step solutions

1. Let u = x and dv = e^x\,dx, so du = dx and v = e^x:

2. Let u = x and dv = \cos x\,dx, so v = \sin x:

3. Let u = \ln x and dv = dx, so du = \frac{1}{x}dx and v = x:

4. Logs come first in LIATE, so u = \ln x and dv = x\,dx, giving v = \frac{x^2}{2}:

5. Let u = x and dv = \sin(2x)\,dx, so v = -\frac{1}{2}\cos(2x):

6. Let u = \arctan x and dv = dx, so du = \frac{1}{1+x^2}dx and v = x:

The last integral is a u-substitution with u = 1 + x^2, giving \frac{1}{2}\ln(1 + x^2).

7. Use the tabular method. Derivatives of x^2: x^2, 2x, 2, 0. Integrals of e^x: e^x, e^x, e^x. Multiply diagonally with signs +, -, +:

8. Tabular again. Derivatives: x^2, 2x, 2, 0. Integrals of \cos x: \sin x, -\cos x, -\sin x. With signs +, -, +:

9. Call the integral I. With u = \sin x and dv = e^x\,dx: I = e^x\sin x - \int e^x\cos x\,dx. Apply parts again with u = \cos x: \int e^x\cos x\,dx = e^x\cos x + I. So:

10. Let I = \int e^{2x}\cos x\,dx. Take u = e^{2x} both times. First, I = e^{2x}\sin x - 2\int e^{2x}\sin x\,dx. Second, \int e^{2x}\sin x\,dx = -e^{2x}\cos x + 2I. Put it together:

11. From problem 1:

12. From problem 4, using \ln e = 1 and \ln 1 = 0:

Common mistakes on these problems

• Choosing u backwards. In problem 4, taking u = x and dv = \ln x\,dx means you must already know \int\ln x\,dx. Follow LIATE.
• Sign errors in the second round. In problems 7 and 8, the minus sign in front of \int v\,du applies to everything that follows. The tabular method keeps the signs straight.
• Switching u and dv types mid-loop. In problem 10, if you take u = e^{2x} first and u = \sin x second, you undo the first step and get I = I.
• Forgetting the \frac{1}{2} in v. When dv = \sin(2x)\,dx, v = -\frac{1}{2}\cos(2x), not -\cos(2x).

Keep practicing

For unlimited problems with new numbers and instant feedback, try the integration by parts lesson and the repeated parts and LIATE lesson. For more examples, see the integral of x e^x, the integral of ln x and the integration by parts guide. Then try trig integral problems.