Calculus · Step-by-step guide

U-Substitution

Use u-substitution to reverse the chain rule in indefinite and definite integrals.

By Tyler Blovat··9 min read
The short answer

Start with the central idea

Choose u as an inner expression whose derivative also appears in the integrand. Replace that derivative-times-dx with du, integrate in u, then substitute back—or change bounds for a definite integral.

01

How it works

Build the method from meaning before memorizing the moves.

These are the decisions that make the procedure reliable. Read them once, then look for each idea in the worked examples below.

1

Look for an inside and its derivative

Typical clues are a power, root, denominator, exponent, or trig input paired with its derivative.

2

Account for constants

Multiply and divide by a constant when the derivative appears up to a constant factor.

3

Finish in one variable

Do not leave a mixture of x and u in the transformed integral.

02

See the structure

A picture makes the relationships easier to remember.

ab
The shaded region represents accumulated signed area between the curve and the x-axis.
03

Worked examples

Follow the reason for each line, then try to reproduce it without looking.

Example 1Power composition

∫2x(x2+1)4dx\displaystyle \int 2x(x^2+1)^4dx.

  1. Let u=x2+1\displaystyle u=x^2+1.
  2. Then du=2x dx\displaystyle du=2x\,dx.
  3. Integrate ∫u4du\displaystyle \int u^4du.
Answer(x2+1)55+C\displaystyle \frac{(x^2+1)^5}{5}+C
Example 2Exponential

∫3e3xdx\displaystyle \int 3e^{3x}dx.

  1. Let u = 3x.
  2. Then du = 3 dx.
  3. The integral becomes ∫eudu\displaystyle \int e^u du.
Answere3x+C\displaystyle e^{3x}+C
Example 3Definite bounds

∫012x(x2+1)2dx\displaystyle \int_0^1 2x(x^2+1)^2dx.

  1. Let u=x2+1\displaystyle u=x^2+1.
  2. Change bounds: x = 0 gives u = 1; x = 1 gives u = 2.
  3. Evaluate ∫12u2du\displaystyle \int_1^2u^2du.
Answer73\displaystyle \frac{7}{3}
04

Common mistakes—and how to avoid them

Accuracy improves fastest when you know what to check.

  • Choosing u without finding du.

  • Leaving x terms after substitution.

  • Changing bounds and then also back-substituting.

05

Turn the explanation into a skill

Reading creates recognition. Independent practice creates recall.

Free printable worksheet · Answer key included

U-Substitution

Practice recognizing reverse-chain-rule patterns.

Preview the worksheet

A strong practice loop: solve one example with the guide open, solve a similar problem without it, explain the method aloud, and return the next day for a short mixed review.

06

Practice problems with worked solutions

Solve each problem on paper, use the hint only if needed, then compare every step.

1

Evaluate ∫2x(x2+1)4 dx\displaystyle \int 2x(x^2+1)^4\,dx.

Show hint

Let u=x2+1\displaystyle u=x^2+1.

Show worked solution
  1. du=2x dx\displaystyle du=2x\,dx.
  2. The integral becomes ∫u4du=u55+C\displaystyle \int u^4du=\frac{u^5}{5}+C.
  3. Substitute back.

Answer: (x2+1)55+C\displaystyle \frac{(x^2+1)^5}{5}+C

2

Evaluate ∫cos⁡(3x) dx\displaystyle \int \cos(3x)\,dx.

Show hint

Let u=3x\displaystyle u=3x.

Show worked solution
  1. du=3dx\displaystyle du=3dx, so dx=du3\displaystyle dx=\frac{du}{3}.
  2. 13∫cos⁡u du\displaystyle \frac{1}{3}\int\cos u\,du.

Answer: 13sin⁡(3x)+C\displaystyle \frac{1}{3}\sin(3x)+C

3

Evaluate ∫xx2+4 dx\displaystyle \int \frac{x}{x^2+4}\,dx.

Show hint

The denominator derivative is 2x.

Show worked solution
  1. Let u=x2+4\displaystyle u=x^2+4, du=2xdx\displaystyle du=2x dx.
  2. The integral is 12∫duu\displaystyle \frac{1}{2}\int \frac{du}{u}.

Answer: 12ln⁡(x2+4)+C\displaystyle \frac{1}{2}\ln(x^2+4)+C

4

Evaluate ∫012xex2 dx\displaystyle \int_0^1 2x e^{x^2}\,dx.

Show hint

Change the bounds after choosing u.

Show worked solution
  1. Let u=x2\displaystyle u=x^2, du=2xdx\displaystyle du=2x dx.
  2. The bounds become 0 and 1.
  3. ∫01eudu=[eu]01\displaystyle \int_0^1e^u du=[e^u]_0^1.

Answer: e−1\displaystyle e-1

5

Evaluate ∫(5x−2)6 dx\displaystyle \int (5x-2)^6\,dx.

Show hint

Let u=5x−2\displaystyle u=5x-2.

Show worked solution
  1. du=5dx\displaystyle du=5dx, so dx=du5\displaystyle dx=\frac{du}{5}.
  2. 15∫u6du=u735\displaystyle \frac{1}{5}\int u^6du=\frac{u^7}{35}.

Answer: (5x−2)735+C\displaystyle \frac{(5x-2)^7}{35}+C

07

Frequently asked questions

Quick answers before you move on.

What should I remember first?

Choose u as an inner expression whose derivative also appears in the integrand. Replace that derivative-times-dx with du, integrate in u, then substitute back—or change bounds for a definite integral.

How do I know whether I understand this topic?

You should be able to name the method, explain why each step is valid, complete a new example without copying, and check whether your answer is reasonable.

What should I do if I keep making the same mistake?

Write the error as a specific checkpoint—for example, “I will identify the hypotenuse before substituting.” Then use that checkpoint on three short problems rather than repeating a full page without feedback.

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