<s>[INST] Trigonometric substitution hint I don't know how to get the following problem started and would appreciate a hint:
$$\int \frac{1}{x^2 \sqrt{9x^2+4}}dx$$
 [/INST] $\int \dfrac{1}{3x^{2}\sqrt{x^{2}+\dfrac{4}{9}}}$
let  $x=\dfrac{2}{3}tan(u)$ so $dx=\dfrac{2}{3}(\sec^2 u)du$
$$I=\int \dfrac{\frac{2}{3}(\sec^{2}u)}{3\frac{4}{9} \tan^2u\sqrt{\frac{4}{9} \tan^2u + \frac{4}{9}}}du$$
$$=\int \frac{\sec^2 u }{\frac{4}{3} \tan^2u \sqrt{\sec^2u}}$$
$$=\frac{3}{4} \csc u\cot u du = -\frac{3}{4} \csc u + c$$
since $x= \frac{2}{3}\tan u$ it follows (draw a triangle) that $\csc(u)=\frac{\sqrt{9x^2+4}}{3x}$ and so the integral reduces to 
$$-\frac{\sqrt{9x^2+4}}{4x} +C$$
</s>