Blog · Calculus guides · 17 October 2026 · 4 min read

Integration by Parts: Formula, Examples & Practice Problems

Q: Is integration by parts the product rule backwards?

Yes. The product rule says (uv)' = u'v + uv'. Integrate both sides and you get uv = ∫ v du + ∫ u dv; move one integral across and that is the formula. So whenever you are unsure of an answer, differentiate it with the product rule and see whether the integrand comes back.

Q: How do you integrate ln x?

Write it as ∫ ln x · 1 dx with u = ln x and dv = dx. Then v = x, and you get x ln x - ∫ 1 dx = x ln x - x + C.

Integration by parts integrates a product of two functions using $\int u d v = uv - \int v d u$ . You pick one factor to differentiate ( $u$ ) and one to integrate ( $d v$ ), so that the new integral is easier than the one you started with. For example, $\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$ .

The formula

\int u d v = uv - \int v d u

It is the product rule run backwards and rearranged. One integral goes in, a slightly different integral comes out, and the whole art is making sure the new one is nicer.

When to use it

Use integration by parts when the integrand is two different kinds of function multiplied together and there is no inside-and-its-derivative pattern for u-substitution to grab: $x sin x$ , $x^{2} e^{3 x}$ , $x ln x$ . It also handles a few lone functions that have no obvious antiderivative, like $ln x$ and $arctan x$ , by treating them as $ln x \cdot 1$ . If substitution works, use substitution: it is quicker. Parts is the tool for when the two factors have nothing to do with each other.

How to do it, step by step

Choose $u$ and $d v$ . Make $u$ the factor that gets simpler when you differentiate it. The LIATE order helps: Logs, Inverse trig, Algebraic (powers of $x$ ), Trig, Exponentials. Whichever comes first in that list is $u$ .
Find $d u$ and $v$ . Differentiate $u$ to get $d u$ , integrate $d v$ to get $v$ . Leave the $+ C$ off $v$ ; one constant at the very end is enough.
Write out $uv - \int v d u$ . Substitute all four pieces into the formula before you do anything clever.
Do the new integral. If it is still a product, you may need parts again. If it looks worse than the original, go back and swap $u$ and $d v$ .
Add $+ C$ and check by differentiating. The product rule should hand you back the integrand exactly.

Worked examples

Example 1: A power times an exponential

Easy

Integrate:

\int x e^{2 x} d x

By LIATE, $x$ (algebraic) beats $e^{2 x}$ (exponential), so $u = x$ and $d v = e^{2 x} d x$ .

u = x, d u = d x, d v = e^{2 x} d x, v = \frac{1}{2} e^{2 x}

Into the formula:

\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \int \frac{1}{2} e^{2 x} d x

The new integral is a plain exponential, which is exactly the point: differentiating $x$ made it vanish.

Answer

\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C

Example 2: When the log has to be u

Medium

Integrate:

\int x ln x d x

Logs come first in LIATE, and for a good reason: you cannot integrate $ln x$ in your head, but you can differentiate it. So $u = ln x$ and $d v = x d x$ .

u = ln x, d u = \frac{1}{x} d x, d v = x d x, v = \frac{x ^{2}}{2}

\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \int \frac{x}{2} d x

The $\frac{1}{x}$ from $d u$ cancels one power of $x$ , and what is left is the power rule in reverse.

Answer

\int x ln x d x = \frac{1}{2} x^{2} ln x - \frac{1}{4} x^{2} + C

Example 3: Parts twice

Harder

Integrate:

\int x^{2} sin x d x

First round: $u = x^{2}$ , $d v = sin x d x$ , so $d u = 2 x d x$ and $v = - cos x$ .

\int x^{2} sin x d x = - x^{2} cos x + \int 2 x cos x d x

The power of $x$ dropped from 2 to 1, so this is working. Second round, on $\int 2 x cos x d x$ : $u = 2 x$ , $d v = cos x d x$ , so $d u = 2 d x$ and $v = sin x$ .

\int 2 x cos x d x = 2 x sin x - \int 2 sin x d x = 2 x sin x + 2 cos x

Put the two rounds together. Watch the signs: the minus in the formula meets the minus from $v = - cos x$ , and that is where most wrong answers to this one come from.

Answer

\int x^{2} sin x d x = - x^{2} cos x + 2 x sin x + 2 cos x + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int x cos (3 x) d x

Problem 2

Medium

Integrate:

\int x^{3} ln x d x

Problem 3

Harder

Integrate:

\int x^{2} e^{- x} d x

Common mistakes

Choosing $u$ backwards. With $\int x e^{2 x} d x$ , setting $u = e^{2 x}$ and $d v = x d x$ gives you $\int \frac{x ^{2}}{2} \cdot 2 e^{2 x} d x$ , which is worse. If the new integral has grown, swap.
Dropping the minus sign. The formula is $uv$ minus $\int v d u$ , and when $v$ is itself negative (as it is for $sin x$ ) the signs stack. Write every line out; this is not the place for mental arithmetic.
Forgetting the chain-rule factor in $v$ . The integral of $e^{2 x}$ is $\frac{1}{2} e^{2 x}$ , not $e^{2 x}$ . Check $v$ by differentiating it before you use it.
Stopping after one round. With $x^{2}$ in front you need two rounds, with $x^{3}$ three. Keep going until the power of $x$ is gone.
Leaving out $+ C$ . One constant, at the end, on every indefinite integral.

Practice it now

Five minutes, no account

Integration by parts is the hardest thing in a Sprint: it only appears at difficulty 6 and 7, the top of Sprint's climb, as $x$ or $x^{2}$ times an exponential, a sine or cosine, or $x^{n} ln x$ . The $+ C$ is optional when you type it: the game accepts any antiderivative.

Play a Sprint See it on the formula sheet

FAQ

How do you choose u in integration by parts?: Use LIATE: Logs, Inverse trig, Algebraic, Trig, Exponentials. The first type in that list becomes $u$ . It is a rule of thumb rather than a law, but it is right for almost every textbook integral.
Why is integration by parts so hard?: Because it is the one rule where you have to make a choice before you can start, and a bad choice only shows itself two lines later. In our answer data it is the least-solved rule in the game: 63% right, with a median of 110 seconds for a correct answer. Slow is normal here.
Is integration by parts the product rule backwards?: Yes. The product rule says $(uv)^{'} = u^{'} v + u v^{'}$ . Integrate both sides and you get $uv = \int v d u + \int u d v$ ; move one integral across and that is the formula. So whenever you are unsure of an answer, differentiate it with the product rule and see whether the integrand comes back.
How do you integrate ln x?: Write it as $\int ln x \cdot 1 d x$ with $u = ln x$ and $d v = d x$ . Then $v = x$ , and you get $x ln x - \int 1 d x = x ln x - x + C$ .
Is there a faster way when I need parts several times?: Yes: tabular integration puts the repeated rounds in a table, so $\int x^{3} e^{x} d x$ takes one line instead of a page.
Where can I find integration by parts practice problems?: Right on this page: "Your turn" has three integration by parts practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

When to use it

Use integration by parts when the integrand is two different kinds of function multiplied together and there is no inside-and-its-derivative pattern for u-substitution to grab:

x sin x

x^{2} e^{3 x}

x ln x

. It also handles a few lone functions that have no obvious antiderivative, like

ln x

and

arctan x

, by treating them as

ln x \cdot 1

. If substitution works, use substitution: it is quicker. Parts is the tool for when the two factors have nothing to do with each other.

How to do it, step by step

Choose $u$ and $d v$ . Make

u

the factor that gets simpler when you differentiate it. The LIATE order helps: Logs, Inverse trig, Algebraic (powers of

x

), Trig, Exponentials. Whichever comes first in that list is

u

Find $d u$ and $v$ . Differentiate

u

to get

d u

, integrate

d v

to get

v

. Leave the

+ C

off

v

; one constant at the very end is enough.

Write out $uv - \int v d u$ . Substitute all four pieces into the formula before you do anything clever.

Do the new integral. If it is still a product, you may need parts again. If it looks worse than the original, go back and swap

u

and

d v

Add $+ C$ and check by differentiating. The product rule should hand you back the integrand exactly.

Worked examples

Example 1: A power times an exponential

Easy

Integrate:

\int x e^{2 x} d x

By LIATE, $x$ (algebraic) beats $e^{2 x}$ (exponential), so $u = x$ and $d v = e^{2 x} d x$ .

u = x, d u = d x, d v = e^{2 x} d x, v = \frac{1}{2} e^{2 x}

Into the formula:

\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \int \frac{1}{2} e^{2 x} d x

The new integral is a plain exponential, which is exactly the point: differentiating $x$ made it vanish.

Answer

\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C

Example 2: When the log has to be u

Medium

Integrate:

\int x ln x d x

Logs come first in LIATE, and for a good reason: you cannot integrate $ln x$ in your head, but you can differentiate it. So $u = ln x$ and $d v = x d x$ .

u = ln x, d u = \frac{1}{x} d x, d v = x d x, v = \frac{x ^{2}}{2}

\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \int \frac{x}{2} d x

The $\frac{1}{x}$ from $d u$ cancels one power of $x$ , and what is left is the power rule in reverse.

Answer

\int x ln x d x = \frac{1}{2} x^{2} ln x - \frac{1}{4} x^{2} + C

Example 3: Parts twice

Harder

Integrate:

\int x^{2} sin x d x

First round: $u = x^{2}$ , $d v = sin x d x$ , so $d u = 2 x d x$ and $v = - cos x$ .

\int x^{2} sin x d x = - x^{2} cos x + \int 2 x cos x d x

The power of $x$ dropped from 2 to 1, so this is working. Second round, on $\int 2 x cos x d x$ : $u = 2 x$ , $d v = cos x d x$ , so $d u = 2 d x$ and $v = sin x$ .

\int 2 x cos x d x = 2 x sin x - \int 2 sin x d x = 2 x sin x + 2 cos x

Put the two rounds together. Watch the signs: the minus in the formula meets the minus from $v = - cos x$ , and that is where most wrong answers to this one come from.

Answer

\int x^{2} sin x d x = - x^{2} cos x + 2 x sin x + 2 cos x + C

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Integrate:

\int x cos (3 x) d x

Problem 2

Medium

Integrate:

\int x^{3} ln x d x

Problem 3

Harder

Integrate:

\int x^{2} e^{- x} d x

Common mistakes

Choosing $u$ backwards. With

\int x e^{2 x} d x

, setting

u = e^{2 x}

and

d v = x d x

gives you

\int \frac{x ^{2}}{2} \cdot 2 e^{2 x} d x

, which is worse. If the new integral has grown, swap.

Dropping the minus sign. The formula is

uv

minus

\int v d u

, and when

v

is itself negative (as it is for

sin x

) the signs stack. Write every line out; this is not the place for mental arithmetic.

Forgetting the chain-rule factor in $v$ . The integral of

e^{2 x}

\frac{1}{2} e^{2 x}

, not

e^{2 x}

. Check

v

by differentiating it before you use it.

Stopping after one round. With

x^{2}

in front you need two rounds, with

x^{3}

three. Keep going until the power of

x

is gone.

Leaving out $+ C$ . One constant, at the end, on every indefinite integral.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

How do you choose u in integration by parts?

Use LIATE: Logs, Inverse trig, Algebraic, Trig, Exponentials. The first type in that list becomes

u

. It is a rule of thumb rather than a law, but it is right for almost every textbook integral.

Why is integration by parts so hard?

Because it is the one rule where you have to make a choice before you can start, and a bad choice only shows itself two lines later. In our answer data it is the least-solved rule in the game: 63% right, with a median of 110 seconds for a correct answer. Slow is normal here.

Is integration by parts the product rule backwards?

Yes. The product rule says

(uv)^{'} = u^{'} v + u v^{'}

. Integrate both sides and you get

uv = \int v d u + \int u d v

; move one integral across and that is the formula. So whenever you are unsure of an answer, differentiate it with the product rule and see whether the integrand comes back.

How do you integrate ln x?

Write it as

\int ln x \cdot 1 d x

with

u = ln x

and

d v = d x

. Then

v = x

, and you get

x ln x - \int 1 d x = x ln x - x + C

Is there a faster way when I need parts several times?

Yes: tabular integration puts the repeated rounds in a table, so

\int x^{3} e^{x} d x

takes one line instead of a page.

Where can I find integration by parts practice problems?

Right on this page: "Your turn" has three integration by parts practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.