Question

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

Answer 1

Let,

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

$\Rightarrow I=\int \frac{xdx}{4\left(\frac{9}{4}-x^2\right)}$

$=\frac{1}{4}\int \frac{xdx}{{\left(\frac{3}{2}\right)}^2-x^2}$

$Putt=x^2$

$\frac{dt}{dx}=2x$

$\frac{1}{2}dt=xdx$

$Now,$

$I=\frac{1}{4}\int \frac{\frac{1}{2}dt}{{\left(\frac{3}{2}\right)}^2-t^2}$

$=\frac{1}{8}\int \frac{1}{\frac{3}{2}-t^2}dt$

$=\frac{1}{8}\times \frac{1}{2\times \frac{3}{2}}\log \left|\frac{\frac{3}{2}+t}{\frac{3}{2}-t}\right|+C$

$=\frac{1}{24}\log \left|\frac{3+2t}{3-2t}\right|+C$

$=\frac{1}{24}\log \left|\frac{3+2x^2}{3-2x^2}\right|+C$