Math Core

Substitution handles integrals built from the chain rule, but it can't touch something as simple-looking as $\displaystyle\int x\cos x\,dx$ or $\displaystyle\int \ln x\,dx$ . Integration by parts is the reverse of the product rule, and it is the main tool for integrating a product of two different kinds of functions.

Where the formula comes from

Start with the product rule for two functions $u$ and $v$ of $x$ :

\frac{d}{dx}\big(uv\big) = u\frac{dv}{dx} + v\frac{du}{dx}.

Integrate both sides and rearrange:

uv = \int u\,dv + \int v\,du \quad\Longrightarrow\quad \int u\,dv = uv - \int v\,du.

Integration by parts

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

For definite integrals:

\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du.

The idea is to split the integrand into two pieces: a piece $u$ that you will differentiate, and a piece $dv$ that you will integrate. The formula trades your original integral for $\displaystyle\int v\,du$ . The method works when that new integral is easier than the one you started with.

Choosing u and dv

The choice is the whole game. You want $u$ to get simpler when you differentiate it, and you need $dv$ to be something you can integrate. A common guide is LIATE: choose $u$ as the first type on this list that appears in the integrand.

Letter	Type	Examples
L	Logarithmic	$\ln x$
I	Inverse trig	$\arctan x$ , $\arcsin x$
A	Algebraic	$x$ , $x^2$ , $3x + 1$
T	Trigonometric	$\sin x$ , $\cos 2x$
E	Exponential	$e^x$ , $e^{-3x}$

Logs and inverse trig functions sit at the top because their derivatives ( $\dfrac{1}{x}$ , $\dfrac{1}{1 + x^2}$ ) are algebraic and much simpler, while their antiderivatives are hard. Exponentials and sines sit at the bottom because they are easy to integrate over and over.

Worked example: A polynomial times a trig function

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

Algebraic comes before trig, so let $u = x$ and $dv = \cos x\,dx$ . Then $du = dx$ and $v = \sin x$ .

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C.

Check: $\dfrac{d}{dx}(x\sin x + \cos x) = \sin x + x\cos x - \sin x = x\cos x$ .

Common mistake

If you choose $u = \cos x$ and $dv = x\,dx$ instead, the formula gives $\tfrac{x^2}{2}\cos x + \displaystyle\int \tfrac{x^2}{2}\sin x\,dx$ , which is harder than the original. When the new integral gets worse, switch your choice of $u$ and $dv$ .

Integrating ln x and arctan x

Some integrals don't look like products at all. Write them as a product with $1$ and let $dv = dx$ .

Worked example: The integral of ln x

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

Let $u = \ln x$ and $dv = dx$ . Then $du = \dfrac{1}{x}\,dx$ and $v = x$ .

\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

The same trick with $u = \arctan x$ gives $\displaystyle\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$ , where the last integral needs a quick substitution.

Definite integrals by parts

With limits, evaluate the $uv$ term at both endpoints and keep the limits on the new integral.

Worked example: A definite integral

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

Let $u = x$ , $dv = e^x\,dx$ , so $du = dx$ and $v = e^x$ .

\int_0^1 x e^x\,dx = \Big[x e^x\Big]_0^1 - \int_0^1 e^x\,dx = (e - 0) - (e - 1) = 1.

Repeated integration by parts

If $u$ is $x^2$ or $x^3$ , one round of parts lowers the power by one, so you apply the formula again. A tabular layout keeps the signs straight. Differentiate $u$ down to zero in one column, integrate $dv$ repeatedly in another, then multiply along the diagonals with alternating signs $+, -, +, \dots$

Worked example: Parts twice, in a table

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

sign	derivatives of $x^2$	integrals of $e^x$
$+$	$x^2$	$e^x$
$-$	$2x$	$e^x$
$+$	$2$	$e^x$
	$0$	$e^x$

Multiply each entry in the middle column by the entry one row down in the right column, using the sign in its row:

\int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x + C = e^x\left(x^2 - 2x + 2\right) + C.

Tip

Sometimes parts brings back the original integral, as in $\displaystyle\int e^x\sin x\,dx$ . After two rounds you get an equation like $I = e^x\sin x - e^x\cos x - I$ . Solve it algebraically: $2I = e^x(\sin x - \cos x)$ , so $I = \tfrac{1}{2}e^x(\sin x - \cos x) + C$ .

Parts with functions given by tables

AP free-response questions often give values of $f$ and $f'$ instead of a formula. Integration by parts still works, because you only need the $uv$ term at the endpoints and the value of the leftover integral. For example, with $u = x$ and $dv = f'(x)\,dx$ :

\int_a^b x f'(x)\,dx = \Big[x f(x)\Big]_a^b - \int_a^b f(x)\,dx.

The last practice problem uses this.

Practice

Practice 1

$\displaystyle\int x e^{2x}\,dx =$

Practice 2

Evaluate $\displaystyle\int_0^{\pi} x\sin x\,dx$ .