Question

Evaluate: $\int \frac{x}{1-x^2}dx$

Answer 1

$I=\int \frac{x}{1-x^2}dx$
$t=1-x^2⟹dt=-2xdx$
$=\int -\frac{1}{2}\frac{1}{t}dt$
$=-\frac{1}{2}\log t$
$=-\frac{1}{2}\log (1-x^2)+c$