Integration by Parts — AP® Calculus BC Notes

Substitution undoes the chain rule. Integration by parts undoes the product rule, and it handles integrals that substitution cannot: $\int x\cos x\,dx$ , $\int xe^{3x}\,dx$ , $\int \ln x\,dx$ . It is a BC-only topic, and it often appears on the free-response section as one part of a question that mixes techniques.

The product rule, run backward

For differentiable functions $u$ and $v$ of $x$ , the product rule says $\frac{d}{dx}(uv)$ $= u\frac{dv}{dx} + v\frac{du}{dx}$ . Integrate both sides:

$uv = \int u\,dv + \int v\,du,$

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

The formula trades one integral, $\int u\,dv$ , for another, $\int v\,du$ . It helps only when the new integral is easier. The figure shows the trade as two areas that together fill a rectangle of area $uv$ .

In the u-v plane, the curve v = ln u runs from (1, 0) to (e, 1). The region under it, from u = 1 to u = e, is shaded violet with area 1 (the integral of v du). The region to its left, from v = 0 to v = 1, is shaded blue with area e − 1, about 1.718 (the integral of u dv). Together they fill the dashed rectangle of area uv = e, so the integral of v du is uv minus the integral of u dv. — **Figure 1.** The two integrals are two areas; together they make the rectangle uv. LevlPrep original diagram.

Choosing u and dv

Split the integrand into $u$ (to be differentiated) and $dv$ (to be integrated, and it must include $dx$ ). Two tests decide the split: $u$ should get simpler when you differentiate it, and you must be able to integrate $dv$ .

Integral	$u$	$dv$	Why
$\int x\cos x\,dx$	$x$	$\cos x\,dx$	$x$ becomes 1
$\int xe^{3x}\,dx$	$x$	$e^{3x}\,dx$	$x$ becomes 1
$\int x^2\ln x\,dx$	$\ln x$	$x^2\,dx$	$\ln x$ becomes $\frac{1}{x}$
$\int \ln x\,dx$	$\ln x$	$dx$	$v = x$
$\int \arctan x\,dx$	$\arctan x$	$dx$	$\arctan x$ becomes $\frac{1}{1 + x^2}$

A useful rule of thumb: logarithms and inverse trigonometric functions go in $u$ ; exponentials, sines and cosines go in $dv$ ; a power of $x$ goes in $u$ unless a logarithm or an inverse trigonometric function is present.

Worked example 1. Find $\int x\cos x\,dx$ .

Label the parts: $u = x$ , $dv = \cos x\,dx$ , so $du = dx$ and $v = \sin x$ .

Apply the formula: $\int x\cos x\,dx$ $= x\sin x - \int \sin x\,dx$ $= x\sin x$ $+ \cos x$ $+ C$ .

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

Writing $u$ , $dv$ , $du$ and $v$ before applying the formula is the habit the scoring guidelines reward. Most lost points come from a wrong $v$ (a sign slip on $\int \sin x\,dx$ ) or from multiplying two antiderivatives, which is never right.

Definite integrals: uv has bounds too

$\int_a^b u\,dv = uv\Big|_a^b - \int_a^b v\,du.$

Worked example 2. Evaluate $\int_1^e x\ln x\,dx$ .

$u = \ln x$ , $dv = x\,dx$ , so $du = \frac{1}{x}\,dx$ and $v = \frac{x^2}{2}$ .

$\int_1^e x\ln x\,dx$ $= \frac{x^2}{2}\ln x\Big|_1^e - \int_1^e \frac{x^2}{2}\cdot\frac{1}{x}\,dx$ $= \frac{e^2}{2} - \frac{x^2}{4}\Big|_1^e$ $= \frac{e^2}{2} - \frac{e^2}{4}$ $+ \frac{1}{4}$ $= \frac{e^2 + 1}{4}.$

Writing $x\ln x - \int_1^e \ldots$ with no bounds on the first term mixes a function with a number; the answer must be a number.

Repeated parts and the tabular method

For $x^2e^{2x}$ , one round of parts leaves $\int xe^{2x}\,dx$ , which needs a second round. The tabular method does every round at once. List $u$ and its derivatives down to 0 in one column, $dv$ and its repeated antiderivatives in the other, and multiply along the diagonals with signs $+, -, +, \ldots$

Worked example 3. Find $\int x^2e^{2x}\,dx$ .

Left column: $x^2$ , $2x$ , $2$ , $0$ . Right column: $e^{2x}$ , $\frac{1}{2}e^{2x}$ , $\frac{1}{4}e^{2x}$ , $\frac{1}{8}e^{2x}$ .

Diagonals: $+x^2\cdot\frac{1}{2}e^{2x} - 2x\cdot\frac{1}{4}e^{2x}$ $+ 2\cdot\frac{1}{8}e^{2x}$ , so

$\int x^2e^{2x}\,dx = e^{2x}\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) + C.$

The table works only when the $u$ column reaches 0, so it suits a power of $x$ times $e^{kx}$ , $\sin kx$ or $\cos kx$ .

When the original integral comes back

For $\int e^x\sin x\,dx$ , two rounds of parts give $\int e^x\sin x\,dx$ $= e^x\sin x - e^x\cos x - \int e^x\sin x\,dx$ . Do not go around a third time. Treat the integral as an unknown $I$ : $2I = e^x(\sin x - \cos x)$ , so $I$ $= \frac{1}{2}e^x(\sin x - \cos x) + C$ .

Parts with functions given by tables

Parts also works when $f$ is known only through values. With $u = x$ and $dv = f'(x)\,dx$ , $\int_a^b xf'(x)\,dx$ $= xf(x)\Big|_a^b - \int_a^b f(x)\,dx$ . You need $f(a)$ , $f(b)$ and the value of $\int_a^b f(x)\,dx$ , which a question gives or lets you find.