Question

$\int \frac{x}{(x-1{)}^2(x+2)}dx$

Answer 1

Solving by the method of partial fractions,

$\frac{x}{{(x-1)}^2(x+2)}=\frac{A}{x-1}+\frac{B}{{(x-1)}^2}+\frac{C}{x+2}$

$x=A(x+2)(x-1)+B(x+2)+C{(x-1)}^2$

Put $x=1\Rightarrow 1=3B$

$\Rightarrow B=\frac{1}{3}$

Put $x=-2\Rightarrow -2=9C$

$\Rightarrow C=\frac{-2}{9}$

Put $x=0\Rightarrow 0=A(0+2)(0-1)+B(0+2)+C{(0-1)}^2$

$\Rightarrow -2A+2B+C=0$

$\Rightarrow -2A+2\times \frac{1}{3}-\frac{2}{9}=0$

$\Rightarrow -2A+\frac{2\times 3-2}{9}=0$

$\Rightarrow -2A+\frac{6-2}{9}=0$

$\Rightarrow -2A=\frac{-4}{9}$

$\Rightarrow A=\frac{-4}{-2\times 9}$

$\therefore A=\frac{2}{9}$

$\therefore A=\frac{2}{9},B=\frac{1}{3},C=\frac{-2}{9}$

$\frac{x}{{(x-1)}^2(x+2)}=\frac{2}{9}\frac{1}{x-1}+\frac{1}{3}\frac{1}{{(x-1)}^2}-\frac{2}{9}\frac{1}{x+2}$

Integrating both sides w.r.t $x$ we get

$\int \frac{x}{{(x-1)}^2(x+2)}=\frac{2}{9}\int \frac{dx}{x-1}+\frac{1}{3}\int \frac{dx}{{(x-1)}^2}-\frac{2}{9}\int \frac{dx}{x+2}$

$=\frac{2}{9}\ln |x-1|+\frac{1}{3}\frac{{(x-1)}^3}{3}-\frac{2}{9}\ln |x+2|+c$

$=\frac{2}{9}[\ln |x-1|-\ln |x+2|]+\frac{1}{3}\frac{{(x-1)}^3}{3}+c$

$=\frac{2}{9}\ln \left(\frac{|x-1|}{|x+1|}\right)+\frac{{(x-1)}^3}{9}+c$