Answer
$$\int \frac{dx}{4x^2+9}=\frac{1}{6} \tan^{-1 }(\frac{2}{3}x)+c.$$
Work Step by Step
Let $ u=\frac{2}{3}x $, then $ du= \frac{2}{3}dx $ and hence
$$\int \frac{dx}{4x^2+9}=\frac{3}{2}\int \frac{du}{9u^2+9} =\frac{3}{2} \frac{1}{9}\int \frac{du}{u^2+1}\\
=\frac{1}{6} \tan^{-1 }u+c=\frac{1}{6} \tan^{-1 }(\frac{2}{3}x)+c.$$
