2. Let $t = x^2$ to get:
$\displaystyle \int \frac{x}{36+(x^2)^2}\;{dx} = \frac{1}{2}\int\frac{1}{6^2+t^2}\;{dt} = \frac{1}{12}\arctan\frac{t}{6}+k = \frac{1}{12}\arctan\frac{x^2}{6}+k$.