Question

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

Answer 1

Using partial fraction

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