Integration by parts

Integration by parts is the product rule used backwards¹. We check it with $\int xe^{-x}\,dx$ .

The formula

\int f g'\,dx = fg - \int f' g\,dx

It comes from integrating both sides of the product rule $(fg)' = f'g + fg'$ and rearranging. One chooses $f$ and $g'$ so that the integral left on the right is simpler than the original.

Working it out

Taking $f = x$ and $g' = e^{-x}$ gives $f' = 1$ and $g = -e^{-x}$ , so the result is as follows.

\int xe^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -(x + 1)e^{-x} + C

Differentiating to check gives $-e^{-x} + (x + 1)e^{-x} = xe^{-x}$ , returning the original.

Making the choice

Choice	Integral left over	Outcome
$f = x$ , $g' = e^{-x}$	$\int e^{-x}dx$	simpler
$f = e^{-x}$ , $g' = x$	$\int \dfrac{x^2}{2}e^{-x}dx$	the degree goes up

The knack is to put the factor that gets simpler under differentiation into $f$ .

It works over an infinite interval

Since $-(x + 1)e^{-x} \to 0$ as $x \to \infty$ , the following holds.

\int_0^{\infty} xe^{-x}\,dx = 0 - (-1) = 1

That says the mean of the exponential distribution is $1$ .

Reading in a factor of $1$

For $\int \log x\,dx$ , taking $f = \log x$ and $g' = 1$ gives $x\log x - \int 1\,dx = x\log x - x$ . Even when nothing looks like a product, reading in a factor of $1$ lets integration by parts apply.

The type that comes back after two rounds

In $\int e^x \sin x\,dx$ , two rounds of integration by parts bring the original integral back on the right. Solving the result as an equation gives the following.

\int e^x \sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C

It looks like going in circles, yet it is a rare type that solves itself.

The humped curve on the graph is $y = xe^{-x}$ , the curve rising from below toward $0$ is the antiderivative $y = -(x + 1)e^{-x}$ , and the large dots are the local maximum and $(0, -1)$ .

Integration by parts, Wikipedia