Question

The value of the integral $\int \frac{2}{(1-x)(1+x^2)}dx$ is (where $C$ is integration constant)

Answer 1

Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{(1-x)}+\frac{Bx+C}{(1+x^2)}$
$2=A(1+x^2)+(Bx+C)(1-x)$
$2=A+A{x}^2+Bx-B{x}^2+C-Cx$
Equating the coefficient of $x^2,x$ and constant term, we obtain
$A-B=0$
$B-C=0$
$A+C=2$
On solving these equations, we obtain
$A=1,B=1$ and $C=1$
$\therefore \frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}$
$=\int \frac{2}{(1-x)(1+x^2)}dx=\int \frac{1}{1-x}dx+\int \frac{x}{1+x^2}dx+\int \frac{1}{1+x^2}dx$
$=-\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{2x}{1+x^2}dx+\int \frac{1}{1+x^2}dx$
$=-\log |x-1|+\frac{1}{2}\log |1+x^2|+{\tan }^{-1}x+C$