U-Substitution Practice Problems with Answers and Solutions

These u-substitution practice problems start with clear inside functions and build up to logs, back-substitution and definite integrals with new limits. Try each one first. The answers come first so you can check quickly, then the full solutions.

The method:

Set u equal to the inside function, find du, replace everything, integrate, then put x back. For a refresher, read the u-substitution rule.

Problems

The derivative is right there

1. \displaystyle\int 2x(x^2 + 1)^5\,dx
2. \displaystyle\int \cos(3x)\,dx
3. \displaystyle\int x\,e^{x^2}\,dx
4. \displaystyle\int x^2\sqrt{x^3 + 1}\,dx

Logs and trig

5. \displaystyle\int \frac{\ln x}{x}\,dx
6. \displaystyle\int \tan x\,dx
7. \displaystyle\int \sin x\cos^4 x\,dx
8. \displaystyle\int \frac{e^x}{1 + e^x}\,dx
9. \displaystyle\int \frac{1}{x\ln x}\,dx

Back-substitution

10. \displaystyle\int x\sqrt{x + 1}\,dx

Definite integrals

11. \displaystyle\int_0^2 x(x^2 + 1)^3\,dx
12. \displaystyle\int_0^{\pi/2}\cos x\,e^{\sin x}\,dx

Answers

Step-by-step solutions

1. Let u = x^2 + 1, so du = 2x\,dx. That is exactly what is in front:

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

3. Let u = x^2, so du = 2x\,dx and x\,dx = \frac{1}{2}du:

4. Let u = x^3 + 1, so du = 3x^2\,dx and x^2\,dx = \frac{1}{3}du:

5. Let u = \ln x, so du = \frac{1}{x}dx:

6. Write \tan x = \frac{\sin x}{\cos x}. Let u = \cos x, so du = -\sin x\,dx:

This is the same as \ln|\sec x| + C. See the integral of tan x.

7. Let u = \cos x, so \sin x\,dx = -du:

8. Let u = 1 + e^x, so du = e^x\,dx. Since 1 + e^x > 0, no absolute value is needed:

9. Let u = \ln x, so du = \frac{1}{x}dx. The integral becomes \int\frac{1}{u}\,du:

10. Let u = x + 1, so du = dx. There is still an x left, so solve for it: x = u - 1. Then:

Put u = x + 1 back to get the answer in the table.

11. Let u = x^2 + 1, so x\,dx = \frac{1}{2}du. Change the limits: x = 0 gives u = 1 and x = 2 gives u = 5:

12. Let u = \sin x, so du = \cos x\,dx. The limits become u = 0 and u = 1:

Common mistakes on these problems

• Leaving dx in the integral. After substituting, everything must be in terms of u and du. A leftover dx means a step is missing.
• Dividing by a variable. In \int e^{x^2}\,dx there is no x factor, so u = x^2 does not work. You can fix constants, not variables.
• Keeping the old limits. In problem 11, integrating u^3 from 0 to 2 gives the wrong answer. Change the limits or switch back to x first.
• Forgetting to switch back. An indefinite answer must be in terms of x.

Keep practicing

For unlimited problems with new numbers and instant feedback, try the u-substitution lesson and the Calculus 2 refresher. For more worked examples, read the u-substitution guide. When substitution stops working on products, move on to integration by parts problems.