Question

Evaluate $\int \frac{3x+1}{\left(x^2-1\right)\left(x-1\right)}dx$

Answer 1

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

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

Consider $\frac{3x+1}{{(x-1)}^2(x+1)}=\frac{A}{x-1}+\frac{B}{{(x-1)}^2}+\frac{C}{x+1}$

$\Rightarrow 3x+1=A(x-1)(x+1)+B(x+1)+C{(x-1)}^2$

Put $x=1\Rightarrow 4=2B\Rightarrow B=2$

Put $x=-1\Rightarrow -3+1=4C\Rightarrow C=-\frac{2}{4}=-\frac{1}{2}$

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

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

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

Integrating both sides, we get

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

$=\frac{1}{2}\log |x-1|-\frac{2}{x-1}-\frac{1}{2}\log |x+1|+c$

$=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}+c$