Integration using partial fractionsAQA A-Level Maths: Revision notes

A fraction such as $\frac{5x+1}{(x-1)(x+2)}$ cannot be integrated directly: the denominator is a product of linear factors. Partial fractions rewrite it as a sum of simpler fractions, each with one linear denominator, which integrate to logarithms. Here the numerator has a lower degree than the denominator and the factors are different. The form is $\frac{5x+1}{(x-1)(x+2)}\equiv\frac{A}{x-1}+\frac{B}{x+2}.$

Multiply through by the common denominator: $5x+1\equiv A(x+2)+B(x-1)$. This is an identity, true for every $x$, so either substitute convenient values or compare coefficients.

• $x=1$: $6=3A$, so $A=2$.
• $x=-2$: $-9=-3B$, so $B=3$. Choosing the value of $x$ that makes a bracket zero removes one unknown at a time. Check by substituting another value, e.g. $x=0$: $\frac{1}{-2}=\frac{2}{-1}+\frac32=-\frac12$.

Each part has the form $\frac{A}{ax+b}$ and integrates to a logarithm: $\int\frac{A}{ax+b}\,dx=\frac Aa\ln|ax+b|+c.$ So $\int\left(\frac{2}{x-1}+\frac{3}{x+2}\right)dx=2\ln|x-1|+3\ln|x+2|+c$. When the coefficient of $x$ is not 1, divide by it: $\int\frac{2}{2x-1}dx=\ln|2x-1|+c$ and $\int\frac{2}{3x+1}dx=\frac23\ln|3x+1|+c$.

Use $\ln a-\ln b=\ln\frac ab$ and $n\ln a=\ln a^n$ to tidy results. With $\frac{4}{(2x-1)(2x+1)}\equiv\frac{2}{2x-1}-\frac{2}{2x+1}$: $\int=\ln|2x-1|-\ln|2x+1|=\ln\left|\frac{2x-1}{2x+1}\right|+c$. For limits: $\int_1^2=\ln\frac35-\ln\frac13=\ln\frac95$. Exact answers are left in logarithms such as $5\ln3-3\ln2$; a decimal is only given when asked.

For $y=\frac{12}{(x+1)(x+3)}\equiv\frac{6}{x+1}-\frac{6}{x+3}$ the area from $x=0$ to $x=3$ is $\left[6\ln\frac{x+1}{x+3}\right]_0^3=6\ln\frac46+6\ln3=6\ln2$. To divide this area equally, find $k$ with $6\ln\frac{3(k+1)}{k+3}=3\ln2$: this gives $\ln\frac{3(k+1)}{k+3}=\frac12\ln2$, so $\frac{3(k+1)}{k+3}=\sqrt2$ and $k=\frac{6\sqrt2-3}{7}$. Remove a logarithm by writing both sides as logs, or by using $e^{\ln a}=a$, and check the answer lies in the stated range.