Guide

Integration By Parts: How And When To Use It

Was this lesson helpful?

Dive Deeper

Generate a follow-up sub-lesson on any aspect of this topic

Guide complete

Integration By Parts: How And When To Use It

Learn to turn hard product integrals into easier ones by reversing the product rule. You will know what to pick for $u$ and $dv$ , spot when the method simplifies, and handle repeat and loop cases without getting stuck.

You are staring at an integral like $\int x e^x\,dx$ and you know how to integrate $e^x$ and how to handle $x$ by itself, but the product blocks you. Integration by parts is the standard move when an integrand is built from two pieces and differentiating one of them makes it simpler. The trick is to trade one derivative for one integral, so the new integral is easier than the original, not just different.

Reverse the product rule

The core idea is that differentiation has a product rule, so integration gets a matching undo button. If you can spot your integrand as one function times the derivative of another, you can rearrange it into a boundary term minus a new integral.

Explore how the product rule turns into the integration-by-parts template.

A clean way to say it is this. You choose $u$ to be the part you want to differentiate, and $dv$ to be the part you want to integrate. After that choice, everything is bookkeeping.

Choose to simplify
If differentiating your $u$ does not get simpler, integration by parts is usually the wrong tool.

Where the formula comes from

Start with the product rule in derivative form:

(uv)'=u'v+uv'.

Integrate both sides with respect to $x$ :

\int(uv)'\,dx=\int u'v\,dx+\int uv'\,dx.

The left side collapses to $uv$ . Then isolate the term you want, usually $\int uv'\,dx$ :

\int uv'\,dx=uv-\int vu'\,dx.

See the same relationship written in differential language and in integral language.

Two small translation notes keep mistakes away.

$dv$ means the whole leftover differential piece, like $e^x\,dx$ or $\cos x\,dx$
$v$ is any antiderivative of $dv$ , so constants belong in $v$ only when they matter later

When integration by parts is the right choice

Integration by parts is most natural for products where one factor gets simpler when differentiated and the other is easy to integrate. Common green lights are:

polynomial times exponential or trig, like $x^2 e^x$ or $x\sin x$
log times something, like $\ln x$ times a polynomial
inverse trig times something integrable, like $\arctan x\cdot x$

A popular selection rule is LIATE, which ranks good candidates for $u$ .

Logarithms
Inverse trig
Algebraic polynomials
Trig
Exponentials

The point is not memorization. It is a bias toward picking $u$ so that $du$ is simpler than $u$ .

Try picking $u$ and watch how the choice changes $du$ and the new integral.

Warning signs you are heading into trouble are when $du$ looks messier than $u$ , or when the new integral resembles the original without being solvable by algebra. That second case is not always bad. It is the loop case you can solve by doing parts twice and then isolating the original integral.

A first complete example $\int x e^x\,dx$

Pick $u=x$ since differentiating kills it down to a constant. Let $dv=e^x\,dx$ since integrating it changes nothing.

Compute the pieces.

$u=x \Rightarrow du=dx$
$dv=e^x\,dx \Rightarrow v=e^x$

Apply the formula $\int u\,dv=uv-\int v\,du$ : [ \int x e^x,dx = x e^x-\int e^x,dx = x e^x-e^x+C. ] A tidy final form is $e^x(x-1)+C$ .

Check yourself by differentiating. $\frac{d}{dx}\bigl(e^x(x-1)\bigr)=e^x(x-1)+e^x=e^x x$ , which matches.

Work through common slip-ups and a quick derivative check on the result.

Parentheses matter
Write $uv-\int v\,du$ as one line before simplifying. Dropping a minus sign usually happens when the new integral is not clearly grouped.

Trig product example $\int x\sin x\,dx$

Use the same strategy. Let $u=x$ so $du=dx$ . Let $dv=\sin x\,dx$ so $v=-\cos x$ .

Now apply integration by parts: [ \int x\sin x,dx = uv-\int v,du = x(-\cos x)-\int(-\cos x),dx = -x\cos x+\int\cos x,dx. ] Finish the remaining integral: [ -x\cos x+\sin x+C. ]

What did simpler mean here. The original integrand was a product. After one round, the new integral was $\int\cos x\,dx$ , a basic single function.

Compare the original integrand to the integrand you get after one step.

Repeated parts and the tabular method

When the polynomial degree is bigger, you often need multiple rounds. For $\int x^2 e^x\,dx$ , choosing $u=x^2$ keeps shrinking when differentiated, and $dv=e^x\,dx$ stays easy to integrate. After enough rounds, the polynomial part becomes $0$ and the process stops.

Doing that with full writing can feel repetitive, so people use the tabular method, which is just integration by parts organized as a table.

Differentiate the polynomial down a column until $0$
Integrate the exponential or trig down the other column
Alternate signs $+,-,+,-,\dots$
Multiply diagonally and add the signed products

See the standard tabular layout and how its diagonal sums match repeated parts.

The tabular method works best when one side eventually becomes $0$ by differentiation, and the other side never gets harder by integrating.

The loop case $\int e^x\sin x\,dx$

Some products do not simplify into a basic integral. They cycle back. A classic is

I=\int e^x\sin x\,dx.

If you do integration by parts once, you get a new integral involving $\int e^x\cos x\,dx$ . Do it again on that new one and $I$ shows up again. That is the moment you stop integrating and start solving algebra.

The pattern is

name the original integral $I$
after two rounds, rewrite the result so it contains $I$ on the right
move that $I$ to the left and factor
divide to solve for $I$

Work through the two-round setup and the algebra step where you isolate the original integral.

Was this lesson helpful?

Dive Deeper

Generate a follow-up sub-lesson on any aspect of this topic

Guide complete

Integration By Parts: How And When To Use It

Learn to turn hard product integrals into easier ones by reversing the product rule. You will know what to pick for $u$ and $dv$ , spot when the method simplifies, and handle repeat and loop cases without getting stuck.

Reverse the product rule

Explore how the product rule turns into the integration-by-parts template.

A clean way to say it is this. You choose $u$ to be the part you want to differentiate, and $dv$ to be the part you want to integrate. After that choice, everything is bookkeeping.

Choose to simplify
If differentiating your $u$ does not get simpler, integration by parts is usually the wrong tool.

Where the formula comes from

Start with the product rule in derivative form:

(uv)'=u'v+uv'.

Integrate both sides with respect to $x$ :

\int(uv)'\,dx=\int u'v\,dx+\int uv'\,dx.

The left side collapses to $uv$ . Then isolate the term you want, usually $\int uv'\,dx$ :

\int uv'\,dx=uv-\int vu'\,dx.

See the same relationship written in differential language and in integral language.

Two small translation notes keep mistakes away.

$dv$ means the whole leftover differential piece, like $e^x\,dx$ or $\cos x\,dx$
$v$ is any antiderivative of $dv$ , so constants belong in $v$ only when they matter later

When integration by parts is the right choice

Integration by parts is most natural for products where one factor gets simpler when differentiated and the other is easy to integrate. Common green lights are:

polynomial times exponential or trig, like $x^2 e^x$ or $x\sin x$
log times something, like $\ln x$ times a polynomial
inverse trig times something integrable, like $\arctan x\cdot x$

A popular selection rule is LIATE, which ranks good candidates for $u$ .

Logarithms
Inverse trig
Algebraic polynomials
Trig
Exponentials

The point is not memorization. It is a bias toward picking $u$ so that $du$ is simpler than $u$ .

Try picking $u$ and watch how the choice changes $du$ and the new integral.

A first complete example $\int x e^x\,dx$

Pick $u=x$ since differentiating kills it down to a constant. Let $dv=e^x\,dx$ since integrating it changes nothing.

Compute the pieces.

$u=x \Rightarrow du=dx$
$dv=e^x\,dx \Rightarrow v=e^x$

Apply the formula $\int u\,dv=uv-\int v\,du$ : [ \int x e^x,dx = x e^x-\int e^x,dx = x e^x-e^x+C. ] A tidy final form is $e^x(x-1)+C$ .

Check yourself by differentiating. $\frac{d}{dx}\bigl(e^x(x-1)\bigr)=e^x(x-1)+e^x=e^x x$ , which matches.

Work through common slip-ups and a quick derivative check on the result.

Parentheses matter
Write $uv-\int v\,du$ as one line before simplifying. Dropping a minus sign usually happens when the new integral is not clearly grouped.

Trig product example $\int x\sin x\,dx$

Use the same strategy. Let $u=x$ so $du=dx$ . Let $dv=\sin x\,dx$ so $v=-\cos x$ .

Now apply integration by parts: [ \int x\sin x,dx = uv-\int v,du = x(-\cos x)-\int(-\cos x),dx = -x\cos x+\int\cos x,dx. ] Finish the remaining integral: [ -x\cos x+\sin x+C. ]

What did simpler mean here. The original integrand was a product. After one round, the new integral was $\int\cos x\,dx$ , a basic single function.

Compare the original integrand to the integrand you get after one step.

Repeated parts and the tabular method

Doing that with full writing can feel repetitive, so people use the tabular method, which is just integration by parts organized as a table.

Differentiate the polynomial down a column until $0$
Integrate the exponential or trig down the other column
Alternate signs $+,-,+,-,\dots$
Multiply diagonally and add the signed products

See the standard tabular layout and how its diagonal sums match repeated parts.

The tabular method works best when one side eventually becomes $0$ by differentiation, and the other side never gets harder by integrating.

The loop case $\int e^x\sin x\,dx$

Some products do not simplify into a basic integral. They cycle back. A classic is

I=\int e^x\sin x\,dx.

The pattern is

name the original integral $I$
after two rounds, rewrite the result so it contains $I$ on the right
move that $I$ to the left and factor
divide to solve for $I$

Work through the two-round setup and the algebra step where you isolate the original integral.

Guide complete

Related content

Integration By Parts: How And When To Use It

Reverse the product rule

Where the formula comes from

When integration by parts is the right choice

A first complete example $\int x e^x\,dx$

Trig product example $\int x\sin x\,dx$

Repeated parts and the tabular method

The loop case $\int e^x\sin x\,dx$

Guide complete

Related content

Integration By Parts: How And When To Use It

Reverse the product rule

Where the formula comes from

When integration by parts is the right choice

A first complete example $\int x e^x\,dx$

Trig product example $\int x\sin x\,dx$

Repeated parts and the tabular method

The loop case $\int e^x\sin x\,dx$

Guide complete

Related content

Integration By Parts: How And When To Use It

Reverse the product rule

Where the formula comes from

When integration by parts is the right choice

A first complete example ∫xex dx\int x e^x\,dx∫xexdx

Trig product example ∫xsin⁡x dx\int x\sin x\,dx∫xsinxdx

Repeated parts and the tabular method

The loop case ∫exsin⁡x dx\int e^x\sin x\,dx∫exsinxdx

Guide complete

Related content

Integration By Parts: How And When To Use It

Reverse the product rule

Where the formula comes from

When integration by parts is the right choice

A first complete example ∫xex dx\int x e^x\,dx∫xexdx

Trig product example ∫xsin⁡x dx\int x\sin x\,dx∫xsinxdx

Repeated parts and the tabular method

The loop case ∫exsin⁡x dx\int e^x\sin x\,dx∫exsinxdx

A first complete example $\int x e^x\,dx$

Trig product example $\int x\sin x\,dx$

The loop case $\int e^x\sin x\,dx$

A first complete example $\int x e^x\,dx$

Trig product example $\int x\sin x\,dx$

The loop case $\int e^x\sin x\,dx$