Integration by Parts

MATIntegration by Parts

Integration by parts is the reverse of the product rule and is used when the integrand is a product of two unlike functions, such as $x e^x$ , $x^2\ln x$ , or $\tan^{-1}x$ . You split the integrand into a part to differentiate ( $u$ ) and a part to integrate ( $dv$ ), choosing $u$ by the ILATE priority so the new integral is simpler.

ISC favours it because it covers logarithmic, inverse-trig, and exponential-times-trig integrals, and it produces the elegant recurring-integral trick for $\int e^x\sin x\,dx$ .

Integration by parts

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

u

is differentiated,

dv

is integrated to give

v

; equivalently

\int u\,v'\,dx = uv - \int u'\,v\,dx

ILATE choice of

u

u:\ \text{I} \to \text{L} \to \text{A} \to \text{T} \to \text{E}

Inverse-trig, Logarithm, Algebraic, Trigonometric, Exponential — whichever comes first is taken as

u

(the function to differentiate).

Single-function trick (log / inverse-trig)

\int \tan^{-1}x\,dx = x\tan^{-1}x - \frac{1}{2}\ln(1+x^2) + C

Write the lone function as

u

and take

dv=1\,dx

v=x

; the same idea gives

\int \ln x\,dx = x\ln x - x + C

Standard exponential result

\int e^{x}\big(f(x)+f'(x)\big)\,dx = e^{x} f(x) + C

Recognise the pattern

f+f'

multiplied by

e^x

; e.g.

\int e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)dx = \dfrac{e^x}{x}+C

with

f(x)=\dfrac{1}{x}

Cyclic (returning) integral

\int e^{ax}\sin bx\,dx = \frac{e^{ax}\big(a\sin bx - b\cos bx\big)}{a^2+b^2} + C

Apply parts twice until the original integral reappears, then solve algebraically for

I

; works for

e^{ax}\cos bx

too.

Pick $u$ as high as possible on the ILATE list so that $du$ is simpler and $dv$ is easy to integrate.
For a single inverse-trig or log function (like $\tan^{-1}x$ or $\ln x$ ), treat it as a product with $1$ : $dv=1\,dx$ .
For $\int x^2 e^x\,dx$ and $\int x^2\ln x\,dx$ , the algebraic power $x^2$ is reduced one degree per application; repeat parts until it disappears.
Cyclic integrals like $\int e^x\sin x\,dx$ require parts twice; do not give up when the original integral returns — move it to the left side and divide.
Memorise the $\int e^x(f+f')\,dx = e^x f + C$ form; spotting it converts a parts problem into instant recognition.
Keep the assignment of $u$ and $dv$ consistent through every step — switching them midway reintroduces the original integral and cancels progress.
For definite integrals, evaluate the $uv$ boundary term as $[uv]_a^b$ and subtract the remaining definite integral.

Where the marks go

Choosing $u$ and $dv$ backwards (e.g. integrating $\ln x$ as $dv$ ) so the new integral is harder than the original.
In the cyclic case, forgetting the sign when the integral reappears, or not dividing by the coefficient after collecting $I$ on one side.
Dropping $+C$ , or for $\int \tan^{-1}x\,dx$ mishandling the second integral $\int\dfrac{x}{1+x^2}\,dx = \dfrac{1}{2}\ln(1+x^2)$ .
Applying parts when a simple substitution would do, wasting steps and inviting algebra errors.

How the board asks it

Numericalilate choice of $u$ and repeated parts
Evaluate $\int x^2 e^{3x}\,dx$ .
Numericaltreating $\ln x$ or $\tan^{-1}x$ as a product with $1$
Evaluate $\int \tan^{-1}x\,dx$ .
Numericalapplying parts twice so the original integral recurs
Evaluate $\int e^{2x}\sin 3x\,dx$ .
Numericalthe $\int e^x\big(f(x)+f'(x)\big)\,dx = e^x f(x) + C$ form
Evaluate $\int e^{x}\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$ .
Numerical $[uv]_a^b$ boundary evaluation
Evaluate $\int_{0}^{1} x\,\tan^{-1}x\,dx$ .

Practise this topic

PreviousIntegration by Partial Fractions NextDefinite Integrals & Evaluation

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.