How should I practice integration by parts?

Study one worked example, cover the solution, reproduce it independently, and then complete mixed practice while checking each result.

Calculus · Step-by-step guide

Integration by Parts

Q: What is the key idea in integration by parts?

Integration by parts reverses the product rule: int u\,dv=uv-int v\,du. Choose u so differentiating simplifies it and choose dv so it can be integrated.

Apply integration by parts, choose u strategically, and handle repeated applications.

By Tyler Blovat·Updated September 22, 2026·9 min read

The short answer

Start with the central idea

Integration by parts reverses the product rule: $\displaystyle \int u\,dv=uv-\int v\,du$ . Choose u so differentiating simplifies it and choose dv so it can be integrated.

\displaystyle \int u\,dv=uv-\int v\,du

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.

Choose u strategically

LIATE—logarithmic, inverse trig, algebraic, trig, exponential—is a useful preference, not an absolute law.

Compute all four pieces

Write u, du, dv, and v before substituting into the formula.

Expect repetition

Polynomial times exponential or trig expressions may require integration by parts more than once.

See the structure

A picture makes the relationships easier to remember.

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

Worked examples

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

Example 1Polynomial times exponential

$\displaystyle \int xe^x dx$ .

Choose u = x and dv = eˣ dx.
Then du = dx and v = eˣ.
Apply uv minus the remaining integral.

Answer $\displaystyle xe^x-e^x+C$

Example 2Logarithm

$\displaystyle \int \ln x\,dx$ .

Treat the integrand as $\displaystyle 1\cdot\ln x$ .
Choose u = ln x and dv = dx.
Then du = dx/x and v = x.

Answer $\displaystyle x\ln x-x+C$

Example 3Trig product

$\displaystyle \int x\cos x\,dx$ .

Choose u = x and dv = cos x dx.
Then du = dx and v = sin x.
Integrate the remaining sine term.

Answer $\displaystyle x\sin x+\cos x+C$

Common mistakes—and how to avoid them

Accuracy improves fastest when you know what to check.

Forgetting the subtraction sign.
Choosing dv that is harder to integrate than the original problem.
Losing +C at the end.

Turn the explanation into a skill

Reading creates recognition. Independent practice creates recall.

Free printable worksheet · Answer key included

Basic Integration

Strengthen the antiderivative fluency used throughout integration by parts.

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.

Practice problems with worked solutions

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

Evaluate $\displaystyle \int xe^x\,dx$ .

Show hint

Choose $\displaystyle u=x$ and $\displaystyle dv=e^xdx$ .

Show worked solution

$\displaystyle du=dx$ and $\displaystyle v=e^x$ .
$\displaystyle uv-\int vdu=xe^x-\int e^xdx$ .

Answer: $\displaystyle xe^x-e^x+C$

Evaluate $\displaystyle \int x\cos x\,dx$ .

Show hint

Choose the algebraic factor as u.

Show worked solution

$\displaystyle u=x,du=dx,dv=\cos xdx,v=\sin x$ .
$\displaystyle x\sin x-\int\sin xdx$ .

Answer: $\displaystyle x\sin x+\cos x+C$

Evaluate $\displaystyle \int \ln x\,dx$ .

Show hint

Treat ln x as u and 1 dx as dv.

Show worked solution

$\displaystyle u=\ln x,du=dx/x,v=x$ .
$\displaystyle x\ln x-\int x(1/x)dx$ .

Answer: $\displaystyle x\ln x-x+C$

Evaluate $\displaystyle \int x^2e^x\,dx$ .

Show hint

Apply integration by parts twice.

Show worked solution

First: $\displaystyle x^2e^x-\int2xe^xdx$ .
For the remaining integral, $\displaystyle \int xe^xdx=xe^x-e^x$ .
Combine terms.

Answer: $\displaystyle e^x(x^2-2x+2)+C$

Evaluate $\displaystyle \int_0^1 xe^x\,dx$ .

Show hint

Use the antiderivative from integration by parts.

Show worked solution

An antiderivative is $\displaystyle e^x(x-1)$ .
Evaluate: $\displaystyle e(0)-1(-1)$ .

Answer: $\displaystyle 1$

Frequently asked questions

Quick answers before you move on.

What should I remember first?

Integration by parts reverses the product rule: $\displaystyle \int u\,dv=uv-\int v\,du$ . Choose u so differentiating simplifies it and choose dv so it can be integrated.

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.

Start with the central idea

How it works

Choose u strategically

Compute all four pieces

Expect repetition

See the structure

Worked examples

∫xexdx\displaystyle \int xe^x dx∫xexdx.

∫ln⁡x dx\displaystyle \int \ln x\,dx∫lnxdx.

∫xcos⁡x dx\displaystyle \int x\cos x\,dx∫xcosxdx.

Common mistakes—and how to avoid them

Turn the explanation into a skill

Basic Integration

Practice problems with worked solutions

Frequently asked questions

U-Substitution

What Is an Integral?

Derivative Rules

$\displaystyle \int xe^x dx$ .

$\displaystyle \int \ln x\,dx$ .

$\displaystyle \int x\cos x\,dx$ .