Integration by parts and partial fractions

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1. Integration by parts

Warm-up

Three quick questions from earlier lessons.

Discover

The product rule says $(x e^x)' = e^x + x e^x$ . Integrate both sides: $x e^x = \displaystyle\int e^x\,dx + \int x e^x\,dx$ . So what is $\displaystyle\int x e^x\,dx$ ?

$x e^x - e^x + C$
$\dfrac{x^2}{2}e^x + C$

Key idea

Trade one integral for an easier one.

Integration by parts:

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

. For definite integrals:

\displaystyle\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du

. Choose

u

so that

du

is simpler.

Worked example

$\displaystyle\int x\sin x\,dx$ .

u = x

du = dx

v = -\cos x

u = x simplifies; dv = sin x dx is easy to integrate.

-x\cos x + \displaystyle\int \cos x\,dx

uv − ∫v du.

-x\cos x + \sin x + C

Finish.

Your turn, guided

Your turn: $\displaystyle\int x e^{2x}\,dx$ , $u = x$ , $v = \tfrac{1}{2}e^{2x}$ .

\tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + C

½xe^{2x} − ∫½e^{2x} dx.

Practice ★

Find $F$ with $F'(x) = 8x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 6x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 7x e^x$ and $F(0) = 0$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_0^{\pi/2} 8x\cos x\,dx$ .
Evaluate $\displaystyle\int_0^{\pi/2} 3x\cos x\,dx$ .

Evaluate $\displaystyle\int_1^{e^{ 3 } } \ln x\,dx$ .

Boss problem. Evaluate $\displaystyle\int_0^{ 2 } x e^x\,dx$ by parts.

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = 8x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 6x e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 7x e^x$ and $F(0) = 0$ .

Summary card

Integration by parts

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

Watch out: the minus sign, and signs inside $\int v\,du$ .

2. Integration by parts: choosing u, repeated parts

Warm-up

Three quick questions from earlier lessons.

Discover

For $\displaystyle\int x\ln x\,dx$ you could take $u = x$ or $u = \ln x$ . With $u = x$ you need $v = \displaystyle\int \ln x\,dx$ first. Which choice is easier?

$u = \ln x$
$u = x$

Key idea

Let $u$ be the factor that simplifies when differentiated.

Good

u

choices, in order: Logarithms, Inverse trig, Algebraic (powers of

x

), Trig, Exponentials (LIATE). With

x^n

times

e^x

\sin x

, apply parts

n

times.

Worked example

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

x^2 e^x - \displaystyle\int 2x e^x\,dx

u = x², v = eˣ.

x^2 e^x - (2xe^x - 2e^x)

Parts again on ∫2xeˣ: 2xeˣ − 2eˣ.

(x^2 - 2x + 2)e^x + C

Tidy.

Your turn, guided

Your turn: $\displaystyle\int x\ln x\,dx$ with $u = \ln x$ , $v = \tfrac{x^2}{2}$ .

\tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C

(x²/2) ln x − ∫x/2 dx.

Practice ★

To integrate $\displaystyle\int x^{ 1 } \ln x\,dx$ by parts, which choice works?
1. $u = \ln x$ , $dv = x^{ 1 }\,dx$
2. $u = x^{ 1 }$ , $dv = \ln x\,dx$
3. $u = x^{ 1 }\ln x$ , $dv = dx$
To integrate $\displaystyle\int x^{ 2 } \ln x\,dx$ by parts, which choice works?
1. $u = \ln x$ , $dv = x^{ 2 }\,dx$
2. $u = x^{ 2 }$ , $dv = \ln x\,dx$
3. $u = x^{ 2 }\ln x$ , $dv = dx$

Evaluate $\displaystyle\int_0^1 x e^{ 2x}\,dx$ .
Evaluate $\displaystyle\int_0^1 x e^{ 5x}\,dx$ .

Watch out!

Practice ★★

Find $F$ with $F'(x) = 5x^2 e^x$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 4x^2 e^x$ and $F(0) = 0$ .

Exit ticket

Three questions to check you have got it.

Evaluate $\displaystyle\int_0^1 x e^{ 2x}\,dx$ .
Evaluate $\displaystyle\int_0^1 x e^{ 5x}\,dx$ .
Evaluate $\displaystyle\int_0^1 x e^{ 3x}\,dx$ .

Summary card

Choosing $u$

LIATE: logs, inverse trig, powers, trig, exponentials.
$x^n e^x$ : parts $n$ times.

Watch out: if the new integral is harder, swap $u$ and $dv$ .

3. Partial fractions

Warm-up

Three quick questions from earlier lessons.

Discover

$\dfrac{1}{x - 1} - \dfrac{1}{x + 1}$ over a common denominator gives $\dfrac{2}{x^2 - 1}$ . So how can you integrate $\dfrac{1}{x^2 - 1}$ ?

Split it as $\tfrac{1}{2}\left(\tfrac{1}{x - 1} - \tfrac{1}{x + 1}\right)$
Use $\arctan x$

Key idea

Split a rational function into simple fractions, then integrate each to a logarithm.

Partial fractions: if

(x - r)(x - s)

with

r \ne s

, then

\dfrac{p(x)}{(x - r)(x - s)} = \dfrac{A}{x - r} + \dfrac{B}{x - s}

(degree of

p

less than 2). Find

A

by multiplying by

x - r

and setting

x = r

; likewise

B

Worked example

$\displaystyle\int \dfrac{5}{(x - 2)(x + 3)}\,dx$ .

A = 1

B = -1

x = 2: A = 5/5; x = −3: B = 5/(−5).

\ln|x - 2| - \ln|x + 3| + C

Integrate each.

\ln\left|\dfrac{x - 2}{x + 3}\right| + C

One log.

Your turn, guided

Your turn: $\dfrac{3}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ .

A = 1

x = 1: 3/(1 + 2).

Practice ★

$\dfrac{ 8 }{(x - 4)(x - 6)} = \dfrac{A}{x - 4} + \dfrac{B}{x - 6}$ . Find $A$ .
$\dfrac{ 5 }{(x - 1)(x - 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x - 3}$ . Find $A$ .
$\dfrac{ 8 }{(x - 1)(x - 5)} = \dfrac{A}{x - 1} + \dfrac{B}{x - 5}$ . Find $A$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_1^2 \dfrac{dx}{x(x + 3)}$ .
Evaluate $\displaystyle\int_1^2 \dfrac{dx}{x(x + 4)}$ .

Evaluate $\displaystyle\int_{ 5 }^{ 6 } \dfrac{dx}{x^2 - 1}$ .

Exit ticket

Three questions to check you have got it.

$\dfrac{ 8 }{(x - 4)(x - 6)} = \dfrac{A}{x - 4} + \dfrac{B}{x - 6}$ . Find $A$ .
$\dfrac{ 5 }{(x - 1)(x - 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x - 3}$ . Find $A$ .
$\dfrac{ 8 }{(x - 1)(x - 5)} = \dfrac{A}{x - 1} + \dfrac{B}{x - 5}$ . Find $A$ .

Summary card

Partial fractions

Factor the denominator.
Find $A$ , $B$ by cover-up.
Integrate: $\displaystyle\int \dfrac{A}{x - r}\,dx = A\ln|x - r|$ .

Watch out: check by recombining the fractions.

4. Choosing an integration technique

Warm-up

Three quick questions from earlier lessons.

Discover

$\displaystyle\int x\cos(x^2)\,dx$ and $\displaystyle\int x\cos x\,dx$ look alike. Which one is a substitution?

$\displaystyle\int x\cos(x^2)\,dx$
$\displaystyle\int x\cos x\,dx$

Key idea

Read the integrand before you start.

Substitution: an inside function with its derivative (up to a constant) as a factor. Parts: a product of unrelated types, such as a power times

e^x

\sin x

\ln x

. Partial fractions: a rational function whose denominator factors. Algebra first: divide, or complete the square.

Worked example

Choose a method for three integrals.

\displaystyle\int 2x(x^2 + 1)^5\,dx

: substitution

x² + 1 inside, 2x outside.

\displaystyle\int x e^{-x}\,dx

: parts

Power times exponential.

\displaystyle\int \dfrac{dx}{x^2 - 4}

: partial fractions

x² − 4 = (x − 2)(x + 2).

Your turn, guided

Your turn: $\displaystyle\int \dfrac{2x}{x^2 + 9}\,dx$ is a substitution with $u = x^2 + 9$ .

\ln(x^2 + 9) + C

∫du/u.

Practice ★

Which method suits $\displaystyle\int x e^{ 2x^2}\,dx$ best?
1. Substitution, $u = 2x^2$
2. Integration by parts
3. Partial fractions

Which method suits $\displaystyle\int x e^{ 6x}\,dx$ best?
1. Integration by parts, $u = x$
2. Substitution, $u = 6x$
3. Partial fractions

Watch out!

Practice ★★

Which method suits $\displaystyle\int \dfrac{dx}{x^2 - 36}$ best?
1. Partial fractions: $x^2 - 36 = (x - 6)(x + 6)$
2. The arctangent formula
3. Integration by parts

Missing exercise F.INT.21-t2

Exit ticket

Three questions to check you have got it.

Which method suits $\displaystyle\int x e^{ 6x}\,dx$ best?
1. Integration by parts, $u = x$
2. Substitution, $u = 6x$
3. Partial fractions
Which method suits $\displaystyle\int x e^{ 3x}\,dx$ best?
1. Integration by parts, $u = x$
2. Substitution, $u = 3x$
3. Partial fractions
Which method suits $\displaystyle\int x e^{ 4x}\,dx$ best?
1. Integration by parts, $u = x$
2. Substitution, $u = 4x$
3. Partial fractions

Summary card

Which technique?

Inside + its derivative: substitution.
Unrelated product: parts.
Factorable denominator: partial fractions.

Watch out: $x^2 + a^2$ (arctan) and $x^2 - a^2$ (fractions) differ.

Answers: Integration by parts and partial fractions

1. Integration by parts

$8 \left(x - 1\right) e^{x} + 8$
$6 \left(x - 1\right) e^{x} + 6$
$7 \left(x - 1\right) e^{x} + 7$

$-8 + 4 \pi$
$-3 + \frac{3 \pi}{2}$

$1 + 2 e^{3}$

$1 + e^{2}$

$8 \left(x - 1\right) e^{x} + 8$
$6 \left(x - 1\right) e^{x} + 6$
$7 \left(x - 1\right) e^{x} + 7$

2. Integration by parts: choosing u, repeated parts

$u = \ln x$ , $dv = x^{ 1 }\,dx$
$u = \ln x$ , $dv = x^{ 2 }\,dx$

$\frac{1}{4} + \frac{e^{2}}{4}$
$\frac{1}{25} + \frac{4 e^{5}}{25}$

$5 \left(x^{2} - 2 x + 2\right) e^{x} - 10$
$4 \left(x^{2} - 2 x + 2\right) e^{x} - 8$

$\frac{1}{4} + \frac{e^{2}}{4}$
$\frac{1}{25} + \frac{4 e^{5}}{25}$
$\frac{1}{9} + \frac{2 e^{3}}{9}$

3. Partial fractions

$-4$
$- \frac{5}{2}$
$-2$

$\frac{\ln{\left(\frac{8}{5} \right)}}{3}$
$\frac{\ln{\left(\frac{5}{3} \right)}}{4}$

$\frac{\ln{\left(\frac{15}{14} \right)}}{2}$

$-4$
$- \frac{5}{2}$
$-2$

4. Choosing an integration technique

Substitution, $u = 2x^2$

Integration by parts, $u = x$

Partial fractions: $x^2 - 36 = (x - 6)(x + 6)$

Integration by parts, $u = x$
Integration by parts, $u = x$
Integration by parts, $u = x$