Integration using partial fractionsEdexcel International A Level Maths: Revision notes

Two results are needed for the rational functions that arise from partial fractions: $\int\frac{1}{ax+b}\,dx=\frac1a\ln|ax+b|+c,\qquad\int\frac{1}{(ax+b)^2}\,dx=-\frac{1}{a(ax+b)}+c.$ Examples: $\int\frac{2}{3x+5}\,dx=\frac23\ln|3x+5|+c$ and $\int\frac{3}{(x-1)^2}\,dx=-\frac{3}{x-1}+c$. The first is a logarithm (power $-1$); the second uses the power rule $\int u^{-2}\,du=-u^{-1}$. Constant multiples stay outside.

A rational expression with a product of linear factors in the denominator cannot be integrated directly. Split it into partial fractions, then integrate each term. For $\frac{4x+5}{(x+1)(x+2)}$ write $\frac{A}{x+1}+\frac{B}{x+2}$, so $4x+5=A(x+2)+B(x+1)$. Substitute $x=-1$: $1=A$. Substitute $x=-2$: $-3=-B$, so $B=3$. $\int\frac{4x+5}{(x+1)(x+2)}\,dx=\int\left(\frac{1}{x+1}+\frac{3}{x+2}\right)dx=\ln|x+1|+3\ln|x+2|+c.$

A squared factor $(x+2)^2$ needs two terms, $\frac{B}{x+2}+\frac{C}{(x+2)^2}$. For $\frac{2x+5}{(x+1)(x+2)^2}$: $\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2},\quad 2x+5=A(x+2)^2+B(x+1)(x+2)+C(x+1).$ $x=-1$ gives $A=3$; $x=-2$ gives $C=-1$; comparing $x^2$ coefficients, $0=A+B$, so $B=-3$. Then $\int\left(\frac{3}{x+1}-\frac{3}{x+2}-\frac{1}{(x+2)^2}\right)dx=3\ln|x+1|-3\ln|x+2|+\frac{1}{x+2}+c.$

Substitute the limits into every term, then combine logarithms using $a\ln x=\ln x^a$, $\ln a+\ln b=\ln ab$ and $\ln a-\ln b=\ln\frac ab$. Example: $\int_0^2\frac{4x+5}{(x+1)(x+2)}\,dx=\left[\ln(x+1)+3\ln(x+2)\right]_0^2=(\ln3+3\ln4)-3\ln2=\ln3+3\ln2=\ln24$. Example: $\int_2^5\frac{5x+7}{(x-1)(x+3)}\,dx=\left[3\ln(x-1)+2\ln(x+3)\right]_2^5=12\ln2-2\ln5=\ln\frac{4096}{25}$. If the integral is indefinite, a modulus sign is needed in $\ln|ax+b|$.

Check partial fractions by substituting a value such as $x=0$ into both sides. Show each integral separately, and keep exact answers: $\ln\frac{4096}{25}$ or $3\ln\frac32-\frac14$ rather than decimals. If 'hence' is used, you must use the partial fractions you found. Curves given by a gradient need the constant $c$ found from a point.