Integration by Parts

Substitution undoes the chain rule. Integration by parts undoes the product rule, and it handles many products that substitution can't, such as a polynomial times an exponential or a logarithm.

The formula

u

and

v

are differentiable functions of

x

, the product rule says

\frac{d}{dx}(uv) = u\,v' + v\,u'

. Integrate both sides and solve for one of the two integrals.

\begin{aligned} uv &= \int u\,v'\,dx + \int v\,u'\,dx \\[4pt] \int u\,v'\,dx &= uv - \int v\,u'\,dx \end{aligned}

The formula trades one integral for another. It pays off when

dv

can be integrated and

\int v\,du

is simpler than the integral you started with.

Choosing u

Choose

u

to be a factor that gets simpler when you differentiate it, and let

dv

be the rest,

dx

included. A common guide ranks the choices for

u

in the order LIATE: logarithms, inverse trigonometric functions, algebraic functions (powers of

x

), trigonometric functions, exponentials. Take

u

from whichever type comes first. It's a guide only: the real test is whether

\int v\,du

is easier.

Example 1

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int (2x - 1)\sin\frac{x}{2}\,dx \\[10pt] &\text{(b)}\;\; \int x^3 \ln(2x)\,dx \end{aligned}

(a) Differentiating

2x - 1

leaves a constant, so take

\begin{aligned} u &= 2x - 1, & dv &= \sin\frac{x}{2}\,dx, \\[4pt] du &= 2\,dx, & v &= -2\cos\frac{x}{2}. \end{aligned}

\begin{aligned} &\int (2x - 1)\sin\frac{x}{2}\,dx \\[4pt] &\quad = -2(2x - 1)\cos\frac{x}{2} \\[4pt] &\qquad\quad + \int 4\cos\frac{x}{2}\,dx \\[4pt] &\quad = -2(2x - 1)\cos\frac{x}{2} \\[4pt] &\qquad\quad + 8\sin\frac{x}{2} + C. \end{aligned}

(b) The logarithm comes first in LIATE, and it has a simple derivative, so take

\begin{aligned} u &= \ln(2x), & dv &= x^3\,dx, \\[4pt] du &= \frac{1}{x}\,dx, & v &= \frac{x^4}{4}. \end{aligned}

\begin{aligned} &\int x^3 \ln(2x)\,dx \\[4pt] &\qquad = \frac{x^4}{4}\ln(2x) - \int \frac{x^3}{4}\,dx \\[4pt] &\qquad = \frac{x^4}{4}\ln(2x) - \frac{x^4}{16} + C. \end{aligned}

A logarithm or an inverse trigonometric function on its own has no second factor. Take

dv = dx

, so

v = x

Example 2

Find the indefinite integral below.

\int \ln\left(x^2 + 1\right)dx

Take

u = \ln(x^2 + 1)

and

dv = dx

, so

du = \frac{2x}{x^2 + 1}\,dx

and

v = x

\begin{aligned} &\int \ln\left(x^2 + 1\right)dx \\[4pt] &\quad = x\ln\left(x^2 + 1\right) - \int \frac{2x^2}{x^2 + 1}\,dx. \end{aligned}

The new integrand's numerator has the same degree as its denominator, so divide:

\frac{2x^2}{x^2 + 1} = 2 - \frac{2}{x^2 + 1}

. Then

\begin{aligned} &\int \ln\left(x^2 + 1\right)dx \\[4pt] &\quad = x\ln\left(x^2 + 1\right) \\[4pt] &\qquad\quad - 2x + 2\arctan x + C. \end{aligned}

Definite integrals and tables

Example 3

Evaluate the integral below.

\int_{\pi/6}^{\pi/2} x\csc^2 x\,dx

Take

u = x

and

dv = \csc^2 x\,dx

, so

du = dx

and

v = -\cot x

. The derivative of

\sin x

\cos x

, so

\int \cot x\,dx = \ln|\sin x| + C

\begin{aligned} &\int_{\pi/6}^{\pi/2} x\csc^2 x\,dx \\[4pt] &\quad = \Big[-x\cot x\Big]_{\pi/6}^{\pi/2} + \int_{\pi/6}^{\pi/2} \cot x\,dx \\[4pt] &\quad = \Big[-x\cot x + \ln|\sin x|\Big]_{\pi/6}^{\pi/2} \\[4pt] &\quad = (0 + 0) - \left(-\frac{\pi\sqrt{3}}{6} + \ln\frac{1}{2}\right) \\[4pt] &\quad = \frac{\pi\sqrt{3}}{6} + \ln 2 \approx 1.600 \end{aligned}

Parts also works when the functions are known only through a table. The product

uv

is evaluated from the table, and the leftover integral is often given.

Example 4

The functions

f

and

g

have continuous derivatives, the table gives some of their values, and

\int_0^2 f'(x)\,g(x)\,dx = -4

. Find

\int_0^2 f(x)\,g'(x)\,dx

$x$	$0$	$2$
$f(x)$	$3$	$-1$
$g(x)$	$2$	$5$

Take

u = f(x)

and

dv = g'(x)\,dx

, so

du = f'(x)\,dx

and

v = g(x)

\begin{aligned} &\int_0^2 f(x)\,g'(x)\,dx \\[4pt] &\quad = \Big[\,f(x)\,g(x)\,\Big]_0^2 \\[4pt] &\qquad\quad - \int_0^2 f'(x)\,g(x)\,dx \\[4pt] &\quad = \big((-1)(5) - (3)(2)\big) - (-4) \\[4pt] &\quad = -7. \end{aligned}

Repeated parts and the tabular method

With

u = x^2

, one round of parts leaves an integral with

x

in place of

x^2

. A second round finishes it.

Example 5

Find the indefinite integral below.

\int \left(x^2 + 1\right)e^{2x}\,dx

First round:

u = x^2 + 1

dv = e^{2x}\,dx

, so

du = 2x\,dx

and

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

\begin{aligned} &\int \left(x^2 + 1\right)e^{2x}\,dx \\[4pt] &\qquad = \frac{1}{2}\left(x^2 + 1\right)e^{2x} - \int x\,e^{2x}\,dx \end{aligned}

Second round, on

\int x\,e^{2x}\,dx

u = x

dv = e^{2x}\,dx

\begin{aligned} \int x\,e^{2x}\,dx &= \frac{1}{2}x\,e^{2x} - \int \frac{1}{2}e^{2x}\,dx \\[4pt] &= \frac{1}{2}x\,e^{2x} - \frac{1}{4}e^{2x} + C \end{aligned}

Put the two rounds together.

\begin{aligned} &\int \left(x^2 + 1\right)e^{2x}\,dx \\[4pt] &\qquad = e^{2x}\left(\frac{x^2 + 1}{2} - \frac{x}{2} + \frac{1}{4}\right) + C \\[4pt] &\qquad = e^{2x}\left(\frac{x^2}{2} - \frac{x}{2} + \frac{3}{4}\right) + C. \end{aligned}

The tabular method organizes the same work. List

u

and its derivatives in one column until a derivative is 0, and

dv

and its antiderivatives in the next. Multiply along each diagonal and attach the signs

+, -, +, \dots

in turn. It finishes the integral only when the

u

column reaches 0.

Example 6

Redo Example 5 with the tabular method.

Sign	$u$ and its derivatives	$dv$ and its antiderivatives
$+$	$x^2 + 1$	$e^{2x}$
$-$	$2x$	$\frac{1}{2}e^{2x}$
$+$	$2$	$\frac{1}{4}e^{2x}$
	$0$	$\frac{1}{8}e^{2x}$

Each entry in the

u

column is multiplied by the entry one row down in the other column:

\begin{aligned} &\left(x^2 + 1\right)\frac{e^{2x}}{2} - 2x\,\frac{e^{2x}}{4} + 2\,\frac{e^{2x}}{8} + C \\[4pt] &\qquad = e^{2x}\left(\frac{x^2}{2} - \frac{x}{2} + \frac{3}{4}\right) + C, \end{aligned}

the same answer as Example 5.

An integral that comes back

For an exponential times a sine or cosine, neither factor ever differentiates to 0, so the tabular method never stops. Integrate by parts twice, choosing the same kind of

u

both times. The original integral returns, and you solve for it.

Example 7

Find the indefinite integral below.

I = \int e^{x/2}\sin 2x\,dx

First round:

u = \sin 2x

dv = e^{x/2}\,dx

, so

du = 2\cos 2x\,dx

and

v = 2e^{x/2}

I = 2e^{x/2}\sin 2x - 4\int e^{x/2}\cos 2x\,dx

Second round, on the new integral:

u = \cos 2x

dv = e^{x/2}\,dx

, so

du = -2\sin 2x\,dx

and

v = 2e^{x/2}

\int e^{x/2}\cos 2x\,dx = 2e^{x/2}\cos 2x + 4I

Substitute that into the first line, and collect the copies of

I

\begin{aligned} I &= 2e^{x/2}\sin 2x \\[4pt] &\qquad - 8e^{x/2}\cos 2x - 16I \\[4pt] 17I &= e^{x/2}\left(2\sin 2x - 8\cos 2x\right) \\[4pt] I &= \frac{2e^{x/2}}{17}\left(\sin 2x - 4\cos 2x\right) + C \end{aligned}

Substituting first

Sometimes a substitution turns the integral into one that parts can handle.