Question

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

Answer 1

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

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

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

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

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

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

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

$=\frac{1}{2}\times \left[\ln \right(x-1)+\ln (x+3)+c]$, where $c$ is a constant

$=\ln \left(\sqrt{(x-1)(x+3)}\right)+\frac{c}{2}$, where $c$ is a constant