# U-Substitution Practice Problems with Answers and Solutions

> 12 u-substitution practice problems with answers and full step-by-step solutions, from (x²+1)⁵ to ln x / x, back-substitution and definite integrals.

Source: https://practicecalculus.com/practice-problems/u-substitution
Updated: 2026-11-06

$$ \int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \quad u = g(x) $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1, Calculus 2 and AP Calculus AB, Unit 6
- **You need:** Basic antiderivatives and the [chain rule](/rules/chain-rule)
- **Main idea:** Pick the inside function as $u$, replace $g'(x)\,dx$ with $du$, and integrate in $u$
- **Practice with new numbers:** [U-substitution lesson](/learn/calculus-1/integrals/u-substitution)

## Key takeaways

- Choose $u$ as the inside of a composite function whose derivative also appears, up to a constant.
- Every $x$ must be gone after the substitution. If one is left, solve $u = g(x)$ for $x$ and replace it.
- In a definite integral, change the limits to $u$-values so you never have to switch back.

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:

$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \qquad u = g(x),\ du = g'(x)\,dx$$

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

| Problem | Answer |
| --- | --- |
| 1 | $\dfrac{(x^2 + 1)^6}{6} + C$ |
| 2 | $\dfrac{\sin(3x)}{3} + C$ |
| 3 | $\dfrac{1}{2}e^{x^2} + C$ |
| 4 | $\dfrac{2}{9}(x^3 + 1)^{3/2} + C$ |
| 5 | $\dfrac{(\ln x)^2}{2} + C$ |
| 6 | $-\ln\lvert\cos x\rvert + C$ |
| 7 | $-\dfrac{\cos^5 x}{5} + C$ |
| 8 | $\ln(1 + e^x) + C$ |
| 9 | $\ln\lvert\ln x\rvert + C$ |
| 10 | $\dfrac{2}{5}(x+1)^{5/2} - \dfrac{2}{3}(x+1)^{3/2} + C$ |
| 11 | $78$ |
| 12 | $e - 1$ |

## Step-by-step solutions

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

$$\int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$$

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

$$\frac{1}{3}\int\cos u\,du = \frac{\sin(3x)}{3} + C$$

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

$$\frac{1}{2}\int e^u\,du = \frac{1}{2}e^{x^2} + C$$

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

$$\frac{1}{3}\int u^{1/2}\,du = \frac{1}{3}\cdot\frac{2}{3}u^{3/2} + C = \frac{2}{9}(x^3+1)^{3/2} + C$$

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

$$\int u\,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$$

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

$$-\int\frac{1}{u}\,du = -\ln|\cos x| + C$$

This is the same as $\ln|\sec x| + C$. See the [integral of tan x](/integrals/tan-x).

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

$$-\int u^4\,du = -\frac{u^5}{5} + C = -\frac{\cos^5 x}{5} + C$$

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

$$\int\frac{1}{u}\,du = \ln(1 + e^x) + C$$

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

$$\ln|\ln x| + C$$

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

$$\int (u - 1)u^{1/2}\,du = \int\left(u^{3/2} - u^{1/2}\right)du = \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C$$

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 $u = 5$:

$$\frac{1}{2}\int_1^5 u^3\,du = \frac{1}{2}\cdot\frac{5^4 - 1^4}{4} = \frac{624}{8} = 78$$

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

$$\int_0^1 e^u\,du = e - 1$$

## 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](/learn/calculus-1/integrals/u-substitution) and the [Calculus 2 refresher](/learn/calculus-2/integration-techniques/u-substitution-refresher). For more worked examples, read the [u-substitution guide](/blog/u-substitution). When substitution stops working on products, move on to [integration by parts problems](/practice-problems/integration-by-parts).

## Frequently asked questions

### How do you choose u in u-substitution?

Look for a function inside another function, such as the $x^2 + 1$ in $(x^2+1)^5$, whose derivative also appears as a factor. A missing constant is fine, since you can divide by it. If the choice leaves stray $x$ terms you cannot replace, try a different $u$.

### What if the derivative of u is off by a constant?

Divide by the constant. In $\int x\,e^{x^2}\,dx$, $du = 2x\,dx$, so $x\,dx = \frac{1}{2}du$. You can only fix constants this way, not missing variables.

### Is u-substitution the reverse of the chain rule?

Yes. The chain rule gives $\frac{d}{dx}F(g(x)) = F'(g(x))g'(x)$, so integrating $F'(g(x))g'(x)$ gives back $F(g(x)) + C$. Every u-substitution answer can be checked by differentiating with the chain rule.