Integration by Parts

A2 · P3 · 3 min

Integration by parts is the product rule run backwards. Use it when the integrand is a product of two different kinds of function, such as $x e^{x}$ , $x \cos x$ or $\ln x$ on its own, and substitution does not work.

Key result

\int u \,\frac{dv}{dx}\, dx = uv - \int v \,\frac{du}{dx}\, dx

For a definite integral:

\int_a^b u \,\frac{dv}{dx}\, dx = \Big[\,uv\,\Big]_a^b - \int_a^b v \,\frac{du}{dx}\, dx

Choosing $u$ : LATE

Pick $u$ as the factor that becomes simpler when differentiated, in this order of preference:

Logarithms: $\ln x$
Algebra: $x$ , $x^2$ , $3x + 1$
Trigonometry: $\sin x$ , $\cos x$
Exponentials: $e^{x}$ , $e^{2x}$

Whatever is left is $\dfrac{dv}{dx}$ , and you must be able to integrate it.

Tip

If $u = \ln x$ then $\dfrac{du}{dx} = \dfrac{1}{x}$ turns the log into an algebraic term. If $u$ is a power of $x$ , each application reduces the power by one, so $x^2 e^x$ needs parts twice.

Method

Write down $u$ and $\dfrac{dv}{dx}$ , then find $\dfrac{du}{dx}$ and $v$ . Do not add a constant when finding $v$ .
Substitute into the formula.
Integrate the new, simpler integral.
Add $+ c$ for an indefinite integral, or evaluate the limits.

A polynomial times an exponential

Find $\displaystyle\int 2x e^{x} \, dx$ .

Solution

Algebra beats exponential, so $u = 2x$ and $\dfrac{dv}{dx} = e^{x}$ .

\frac{du}{dx} = 2, \qquad v = e^{x}

\int 2x e^{x} \, dx = 2x e^{x} - \int 2 e^{x} \, dx = 2x e^{x} - 2e^{x} + c = 2e^{x}(x - 1) + c

Integrating ln x

Find $\displaystyle\int \ln x \, dx$ .

Solution

There is only one factor, so treat it as $\ln x \times 1$ with $u = \ln x$ and $\dfrac{dv}{dx} = 1$ .

\frac{du}{dx} = \frac{1}{x}, \qquad v = x

\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + c

Parts with a chain-rule integral

Find $\displaystyle\int x \sqrt{1 + x} \, dx$ .

Solution

$u = x$ , $\dfrac{dv}{dx} = (1 + x)^{1/2}$ , so $\dfrac{du}{dx} = 1$ and $v = \dfrac{2}{3}(1 + x)^{3/2}$ .

\int x \sqrt{1 + x} \, dx = \frac{2}{3} x (1 + x)^{3/2} - \int \frac{2}{3}(1 + x)^{3/2} \, dx = \frac{2}{3} x (1 + x)^{3/2} - \frac{4}{15}(1 + x)^{5/2} + c

Factorising, with $(1+x)^{3/2}$ common:

= \frac{2}{15}(1 + x)^{3/2}\big(5x - 2(1 + x)\big) + c = \frac{2}{15}(1 + x)^{3/2}(3x - 2) + c

A definite integral

Evaluate $\displaystyle\int_0^2 x e^{x} \, dx$ , giving the exact answer.

Solution

$u = x$ , $\dfrac{dv}{dx} = e^{x}$ , so $\dfrac{du}{dx} = 1$ and $v = e^{x}$ .

\int_0^2 x e^{x} \, dx = \Big[x e^{x}\Big]_0^2 - \int_0^2 e^{x} \, dx = \big(2e^{2} - 0\big) - \Big[e^{x}\Big]_0^2 = 2e^{2} - (e^{2} - 1) = e^{2} + 1

Numerically this is $8.39$ to 3 s.f.

Applying parts twice

When $u$ is $x^2$ (or the integral is $e^{x}\sin x$ type) one application leaves an integral that still needs parts.

Parts twice

Find $\displaystyle\int x^2 \cos x \, dx$ .

Solution

First pass: $u = x^2$ , $\dfrac{dv}{dx} = \cos x$ , so $\dfrac{du}{dx} = 2x$ , $v = \sin x$ .

\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx

Second pass on $\int 2x \sin x \, dx$ : $u = 2x$ , $\dfrac{dv}{dx} = \sin x$ , so $\dfrac{du}{dx} = 2$ , $v = -\cos x$ .

\int 2x \sin x \, dx = -2x \cos x + \int 2 \cos x \, dx = -2x \cos x + 2 \sin x

Putting it together:

\int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + c

Watch out

The classic slip is a sign error in the second application. Write the second integral out on its own line, finish it, and only then substitute back with brackets around the whole thing.

Exam tip

P3 questions often say "use integration by parts" and then ask for an exact answer. Keep $e$ and $\ln$ terms exact and only convert to decimals if the question asks. Marks are typically 1 for the correct $u$ , $v$ setup, 1 for the substitution, and the rest for accuracy.

Practice

Question

Find $\displaystyle\int x \sin 2x \, dx$ .
Find $\displaystyle\int x^2 \ln x \, dx$ .
Show that $\displaystyle\int_1^{e} \ln x \, dx = 1$ .
Find $\displaystyle\int_0^{\pi/2} x \cos x \, dx$ , giving the exact value.

Answers

$-\dfrac{1}{2} x \cos 2x + \dfrac{1}{4} \sin 2x + c$
$\dfrac{1}{3} x^3 \ln x - \dfrac{1}{9} x^3 + c$
$\big[x \ln x - x\big]_1^{e} = (e - e) - (0 - 1) = 1$
$\dfrac{\pi}{2} - 1$

Choosing uuu: LATE

Method

Applying parts twice

Practice

Console

Choosing $u$ : LATE