Integration by Parts — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

x\sin x

into

\sin x + x\cos x

. Read backward, it says that an integral of a product can be traded for a different integral, often a much easier one. That trade is integration by parts, and it is how you integrate

x\cos x

,

xe^{3x}

and even

\ln x

, none of which substitution can touch.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\frac{d}{dx}\left(x\sin x\right) =$

$\sin x + x\cos x$
$\cos x$
$x\cos x$
$\sin x - x\cos x$

Show the answer

Product rule: $(1)\sin x + x\cos x$ .

Correct: $\sin x + x\cos x$ :
$\cos x$ :
$x\cos x$ :
$\sin x - x\cos x$ :

2. $\int e^{3x}\,dx =$

$\frac{1}{3}e^{3x} + C$
$3e^{3x} + C$
$e^{3x} + C$
$\frac{1}{4}e^{4x} + C$

Show the answer

A linear inside function: divide by 3.

Correct: $\frac{1}{3}e^{3x} + C$ :
$3e^{3x} + C$ :
$e^{3x} + C$ :
$\frac{1}{4}e^{4x} + C$ :

Part 4 · See it

See it first

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. — Along v = ln u from (1, 0) to (e, 1), the area under the curve (∫ v du) and the area to its left (∫ u dv) fill the e-by-1 rectangle. One area is the rectangle minus the other: that is integration by parts. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The product rule says $\frac{d}{dx}(uv)$ $= u\frac{dv}{dx} + v\frac{du}{dx}$ integrating both sides gives $uv = \int u\,dv + \int v\,du$
Solve for one of the integrals $\int u\,dv = uv - \int v\,du$ : an integral traded for another
Choose $u$ to get simpler when differentiated and $dv$ to be easy to integratethe new integral $\int v\,du$ is easier than the old one
The new integral still holds a power of $x$ repeat parts, or lay the steps out in a table (the tabular method)
The integral is definiteevaluate $uv$ at both bounds: $\int_a^b u\,dv$ $= uv\Big|_a^b - \int_a^b v\,du$

Part 6 · Key ideas

Key ideas

Formula: $\int u\,dv = uv - \int v\,du$ . Write $u$ , $dv$ , $du$ and $v$ before you start.
Put in $u$ what gets simpler when differentiated (a power of $x$ , $\ln x$ , an inverse trigonometric function); put in $dv$ what you can integrate ( $e^{kx}$ , $\sin kx$ , $\cos kx$ ).
A power $x^n$ times $e^{kx}$ , $\sin kx$ or $\cos kx$ takes $n$ rounds of parts; the tabular method does them all at once with alternating signs.
Definite integrals: $\int_a^b u\,dv$ $= uv\Big|_a^b - \int_a^b v\,du$ . The $uv$ term needs its bounds too.

Part 7 · Misconception

A common mistake

The wrong idea: $\int x\cos x\,dx$ $= \frac{x^2}{2}\sin x + C$ : integrate each factor.

What actually happens: The antiderivative of a product is not the product of the antiderivatives (differentiate $\frac{x^2}{2}\sin x$ to see). With $u = x$ and $dv = \cos x\,dx$ : $x\sin x - \int \sin x\,dx$ $= x\sin x$ $+ \cos x$ $+ C$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Tabular integration of $x^2 e^{2x}$

A student finds

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

by the tabular method. The left column holds

u = x^2

and its derivatives down to 0; the right column holds

dv = e^{2x}\,dx

and its repeated antiderivatives. Each diagonal product, from a row of the left column to the next row of the right column, is one term of the answer, with the signs alternating.

Tabular layout for the integral of x squared times e to the 2x
Sign	$u$ and its derivatives	$dv$ and its antiderivatives
$+$	$x^2$	$e^{2x}$
$-$	$2x$	$\frac{1}{2}e^{2x}$
$+$	$2$	$\frac{1}{4}e^{2x}$
	$0$	$\frac{1}{8}e^{2x}$

1. Why is $x^2$ put in the $u$ column rather than $e^{2x}$ ?

Because $e^{2x}$ has no derivative that can be written down
Because the $u$ column is for the factor with the larger value
Either choice gives the answer in exactly three rows
Its derivatives reach 0, while $e^{2x}$ can be integrated again and again

Show the answer

The table stops when the $u$ column reaches 0. Powers of $x$ do that after a few derivatives; $e^{2x}$ only gets multiplied by constants, so it belongs in the $dv$ column.

Because $e^{2x}$ has no derivative that can be written down: $e^{2x}$ has the derivative $2e^{2x}$ . The reason is what happens to $x^2$ : its derivatives reach 0.
Because the $u$ column is for the factor with the larger value: Size has nothing to do with it. Choose $u$ so that its derivatives get simpler and end at 0.
Either choice gives the answer in exactly three rows: With $u = e^{2x}$ the $dv$ column would integrate $x^2$ upward to $\frac{x^3}{3}, \frac{x^4}{12}, \ldots$ and the table would never end.
Correct: Its derivatives reach 0, while $e^{2x}$ can be integrated again and again: Right: differentiating $x^2$ three times gives 0, which ends the table, and $e^{2x}$ never gets harder to integrate.

2. Which product is the second term of the answer?

$-2x \cdot \frac{1}{4}e^{2x}$
$-2x \cdot \frac{1}{2}e^{2x}$
$+2x \cdot \frac{1}{4}e^{2x}$
$-2x \cdot \frac{1}{8}e^{2x}$

Show the answer

Terms: $+x^2 \cdot \frac{1}{2}e^{2x}$ , then $-2x \cdot \frac{1}{4}e^{2x}$ , then $+2 \cdot \frac{1}{8}e^{2x}$ .

Correct: $-2x \cdot \frac{1}{4}e^{2x}$ : Right: the second row's $u$ entry, $2x$ , times the next row's $dv$ entry, $\frac{1}{4}e^{2x}$ , with the second sign, minus.
$-2x \cdot \frac{1}{2}e^{2x}$ : That multiplies across one row. Each term runs on a diagonal: $2x$ times the entry one row lower, $\frac{1}{4}e^{2x}$ .
$+2x \cdot \frac{1}{4}e^{2x}$ : The diagonal is right but the sign is not: the signs alternate $+, -, +$ , so the second term is subtracted.
$-2x \cdot \frac{1}{8}e^{2x}$ : That skips a row. The diagonal from $2x$ goes to the very next row of the right column, $\frac{1}{4}e^{2x}$ .

3. Use the table to find $\int x^2 e^{2x}\,dx$ . Include $+ C$ .

Type an expression for $\int x^2 e^{2x}\,dx =$ .

Show the answer

$\frac{1}{2}x^2e^{2x} - \frac{1}{2}xe^{2x}$ $+ \frac{1}{4}e^{2x}$ $+ C$ $= e^{2x}\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right)$ $+ C$ . Check by differentiating: the product rule gives back $x^2e^{2x}$ .

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

4. Find $\int x\cos x\,dx$ . Include $+ C$ .

Type an expression for $\int x\cos x\,dx =$ .

Show the answer

$u = x$ , $dv = \cos x\,dx$ , $du = dx$ , $v = \sin x$ : $x\sin x - \int \sin x\,dx$ $= x\sin x$ $+ \cos x$ $+ C$ .

Answer: $\int x\cos x\,dx =$ $x\,\sin(x) + \cos(x) + C$

5. Find $\int \ln x\,dx$ for $x > 0$ . Include $+ C$ .

Type an expression for $\int \ln x\,dx =$ .

Show the answer

$u = \ln x$ , $dv = dx$ , $du = \frac{1}{x}\,dx$ , $v = x$ : $x\ln x - \int x \cdot \frac{1}{x}\,dx$ $= x\ln x - x + C$ .

Answer: $\int \ln x\,dx =$ $x\,\ln(x) - x + C$

6. A student writes $\int_1^2 xe^x\,dx$ $= xe^x - \int_1^2 e^x\,dx$ $= xe^x - (e^2 - e)$ . What is wrong?

$du$ should be $x\,dx$ , not $dx$
The integral of $e^x$ on $[1, 2]$ is $e^2$ , not $e^2 - e$
The term $uv = xe^x$ is never evaluated from 1 to 2
$u$ and $dv$ should be swapped so that $u = e^x$

Show the answer

$xe^x\Big|_1^2 - (e^2 - e)$ $= (2e^2 - e) - (e^2 - e)$ $= e^2$ . Leaving $uv$ unevaluated mixes a function with a number, and the scoring guidelines give no answer point for it.

$du$ should be $x\,dx$ , not $dx$ : With $u = x$ , $du = dx$ . That step is right; $xe^x$ was left without its bounds.
The integral of $e^x$ on $[1, 2]$ is $e^2$ , not $e^2 - e$ : $\int_1^2 e^x\,dx$ $= e^2 - e$ is right. The problem is the unevaluated $xe^x$ .
Correct: The term $uv = xe^x$ is never evaluated from 1 to 2: Right: a definite integral by parts is $uv\Big|_1^2 - \int_1^2 v\,du$ ; the answer must be a number, here $e^2$ .
$u$ and $dv$ should be swapped so that $u = e^x$ : $u = x$ is the right choice: it simplifies to 1. The error is the missing bounds on $xe^x$ .

Part 9 · Summary

Summary

Integration by parts is the product rule run backward:

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

. Choose

u

to simplify when differentiated and

dv

to integrate easily, write out

u

,

dv

,

du

and

v

, and in a definite integral evaluate

uv

at both bounds. Repeated parts, or the tabular method, handles

x^n

times an exponential or a sine.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Tabular integration of x2e2xx^2 e^{2x}

Summary

What comes next

Connections

Tabular integration of $x^2 e^{2x}$