Partial fractionsEdexcel A-Level Maths: Revision notes

A rational function with a factorised denominator can be written as a sum of simpler fractions, called partial fractions. Each linear factor gives one term:

• distinct factors: $\frac{5x+1}{(x-1)(x+2)}\equiv\frac{A}{x-1}+\frac{B}{x+2}$
• three distinct factors: $\frac{\ldots}{(ax+b)(cx+d)(ex+f)}\equiv\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}$
• a repeated factor: $\frac{\ldots}{(ax+b)(cx+d)^2}\equiv\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d)^2}$ The numerator is constant or linear over a quadratic denominator, and the numerator has lower degree than the denominator.

Multiply both sides by the full denominator to get an identity in $x$, then choose a method. Substitution: choose $x$ values that make a bracket zero. For $5x+1\equiv A(x+2)+B(x-1)$: $x=1$ gives $6=3A$, so $A=2$; $x=-2$ gives $-9=-3B$, so $B=3$. Equating coefficients: compare powers of $x$ and solve simultaneously. Here $x$: $5=A+B$, constants: $1=2A-B$, giving $A=2$, $B=3$. For $g(x)=\frac{3x^2+7x+5}{(x+1)(x+2)^2}$, use $3x^2+7x+5\equiv A(x+2)^2+B(x+1)(x+2)+C(x+1)$. $x=-1$ gives $A=1$ and $x=-2$ gives $C=-3$; then $x=0$ gives $5=4+2B-3$, so $B=2$.

Use $\int\frac{1}{ax+b}\,dx=\frac1a\ln|ax+b|+c$ and, for a squared bracket, $\int\frac{1}{(ax+b)^2}\,dx=-\frac{1}{a(ax+b)}+c$. Example: $\int\frac{5x+1}{(x-1)(x+2)}\,dx=\int\left(\frac{2}{x-1}+\frac{3}{x+2}\right)dx=2\ln|x-1|+3\ln|x+2|+c$. Example (repeated factor): $\int_0^1\left(\frac{1}{x+1}+\frac{2}{x+2}-\frac{3}{(x+2)^2}\right)dx=\left[\ln(x+1)+2\ln(x+2)+\frac{3}{x+2}\right]_0^1=2\ln3-\ln2-\frac12$. For $\int\frac{1}{2-x}\,dx$ the coefficient of $x$ is $-1$, so the result is $-\ln|2-x|$.

Writing $f$ as partial fractions avoids the quotient rule. For $f(x)=\frac{7x+4}{(x+1)(2x-1)}=\frac{1}{x+1}+\frac{5}{2x-1}$: $f'(x)=-\frac{1}{(x+1)^2}-\frac{10}{(2x-1)^2},$ so $f'(0)=-1-10=-11$. The factor $2$ in $2x-1$ comes from the chain rule.

Partial fractions turn a hard expansion into binomial expansions of $(1+px)^n$, valid for $|px|<1$. For $f(x)=\frac{5-x}{(1+x)(2-x)}=\frac{2}{1+x}+\frac{1}{2-x}$: $\frac{2}{1+x}=2(1-x+x^2-\ldots)$, valid for $|x|<1$. $\frac{1}{2-x}=\frac12\left(1-\frac x2\right)^{-1}=\frac12\left(1+\frac x2+\frac{x^2}{4}+\ldots\right)$, valid for $|x|<2$. Adding: $f(x)=\frac52-\frac74x+\frac{17}{8}x^2+\ldots$, valid where both hold, $|x|<1$. The same method gives, for $\frac{7x+4}{(x+1)(2x-1)}$, the expansion $-4-11x-19x^2-\ldots$ for $|x|<\frac12$.