Integration by Partial Fractions

MATPartial Fractions

Partial fractions integrate a proper rational function $\dfrac{P(x)}{Q(x)}$ by decomposing it into a sum of simpler fractions, one per factor of the denominator, each of which integrates to a log or arctan. The method depends on the type of factor in $Q(x)$ — distinct linear, repeated linear, or irreducible quadratic — and each type has a fixed template.

ISC examines it because it converts a forbidding fraction into routine standard integrals, and it is the gateway to integrating most rational functions.

Distinct linear factors

\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

a\ne b

; find

A,B

by equating numerators or by the cover-up substitution

x=a

x=b

. Each piece integrates to a

\ln

, e.g.

A\ln|x-a|

Repeated linear factor

\frac{px+q}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}

A factor raised to power

k

contributes terms for every power

1,2,\dots,k

; here

\int\dfrac{B}{(x-a)^2}\,dx = -\dfrac{B}{x-a}+C

Irreducible quadratic factor

\frac{px+q}{(x^2+c)(x-b)} = \frac{Ax+B}{x^2+c} + \frac{C}{x-b}

An irreducible quadratic

x^2+c

(discriminant

<0

) gets a linear numerator

Ax+B

; it integrates to a

\ln

plus an arctan term.

Resulting standard integrals

\int \frac{dx}{x-a} = \ln|x-a| + C, \qquad \int \frac{dx}{x^2+c} = \frac{1}{\sqrt{c}}\tan^{-1}\frac{x}{\sqrt{c}} + C

c>0

; these are the pieces every decomposition reduces to (logs for linear parts, arctan for quadratic parts).

First make the fraction proper: if $\deg P \ge \deg Q$ , divide so that $\dfrac{P}{Q} = (\text{polynomial}) + \dfrac{R}{Q}$ with $\deg R < \deg Q$ .
Factor $Q(x)$ completely; the form of the decomposition is dictated entirely by the factor types (distinct, repeated, quadratic).
The cover-up (Heaviside) method finds the constant over $(x-a)$ quickly: cover that factor and substitute $x=a$ into the rest.
For a repeated factor $(x-a)^k$ you need all $k$ partial fractions $\dfrac{A_1}{x-a}+\dots+\dfrac{A_k}{(x-a)^k}$ , not just one.
After decomposing, integrate term by term: linear factors give $\ln$ , the $\dfrac{1}{(x-a)^2}$ type gives $-\dfrac{1}{x-a}$ , quadratic factors give $\ln$ plus arctan.
Verify the constants by substituting one convenient extra value of $x$ , or by recombining the fractions.
An irreducible quadratic in the denominator (such as $x^2+1$ ) keeps an $Ax+B$ numerator; do not split it into real linear factors.

Where the marks go

Forgetting to do polynomial long division first when the fraction is improper ( $\deg P \ge \deg Q$ ).
Writing only $\dfrac{A}{(x-a)^2}$ for a repeated factor and omitting the $\dfrac{B}{x-a}$ term (or vice-versa).
Giving an irreducible quadratic factor a constant numerator instead of the required linear $Ax+B$ .
Sign and absolute-value errors in the final logs, or dropping $+C$ after integrating each piece.

How the board asks it

Numericaldistinct linear factors
Evaluate $\displaystyle\int \dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx$ .
Numericalrepeated linear factor
Find $\displaystyle\int \dfrac{x+2}{(x-1)^2(x+1)}\,dx$ .
Numericalimproper fraction, divide first
Evaluate $\displaystyle\int \dfrac{x^2}{x^2-5x+6}\,dx$ by first reducing the integrand to a proper fraction.
Numericalirreducible quadratic factor
Find $\displaystyle\int \dfrac{2x}{(x^2+1)(x-1)}\,dx$ .
Numericaldefinite integral via partial fractions
Evaluate $\displaystyle\int_{0}^{1} \dfrac{2x+3}{(x+1)(x^2+4)}\,dx$ .
Numericalsubstitution to a rational function
Evaluate $\displaystyle\int \dfrac{\cos x}{(1+\sin x)(2+\sin x)}\,dx$ by reducing it to partial fractions.

Practise this topic

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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.