Question

Prove that $\int \frac{dx}{(1+x)(1+x^2)}=\frac{1}{2}\log (1+x)+\frac{1}{4}\log (1+x^2)+\frac{1}{2}{\tan }^{-1}x$.

Answer 1

$\int \frac{1}{(x+1)(x^2+1)}dx=\int (\frac{1-x}{2(x^2+1)}+\frac{1}{2(x-1)})dx$
$=\frac{1}{2}\int \frac{1-x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x+1}dx$
$=\frac{1}{2}\int (\frac{1}{x^2+1}-\frac{x}{x^2+1})dx+\frac{1}{2}\int \frac{1}{x+1}dx$
$=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x+1}dx$
$=-\frac{1}{4}\log (x^2+1)+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log (x+1)$