Question

Evaluate $\int \frac{x}{9-4x^2}dx$

Answer 1

Let $t=9-4x^2\Rightarrow dt=8xdx$

$\Rightarrow xdx=\frac{1}{8}dt$

$\int \frac{x}{9-4x^2}dx$

$=\frac{1}{8}\int \frac{dt}{t}$

$=\frac{1}{8}\log t+c$ where $c$ is the constant of integration.

$=\frac{1}{8}\log (9-4x^2)+c$ where $t=9-4x^2$