Question

Integrate the function $\frac{x}{9-4x^2}$

Answer 1

Put $9-4x^2=t$
$\therefore -8xdx=dt$
$\Rightarrow \int \frac{x}{9-4x^2}dx=\frac{-1}{8}\int \frac{1}{t}dt$
$=\frac{-1}{8}\log |t|+C$
$=\frac{-1}{8}\log |9-4x^2|+C$