Trigonometric substitution hint I don't know how to get the following problem started and would appreciate a hint:
$$\int \frac{1}{x^2 \sqrt{9x^2+4}}dx$$
 A: write the integral as : 
$$I=\int \frac{dx}{x^3 \sqrt{9+\dfrac{4} {x^2}}}$$
Now you can subsitute t = $9 + \dfrac{4} {x^2}$
$$
\begin{align}
dt &= -\dfrac{4}{x^3}dx \\
\dfrac{dt}{-8} &= \dfrac{dx}{x^3} \\
I&=\int \frac{dt}{-8 \sqrt{t}} \\
\end{align}
$$ 
This can be easily integrated ! 
In problems like these, sometimes you can get a very neat answer by taking x^n from the root.( I am unable to express it better and if you can write it better, please edit the answer.)
A: $\int \dfrac{1}{3x^{2}\sqrt{x^{2}+\dfrac{4}{9}}}$
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 
$$-\frac{\sqrt{9x^2+4}}{4x} +C$$
A: Jagy's substitution is the best here
substitute $x=\frac 23 \sinh(t)$ and $ dx=\frac 23 \cosh(t)dt$
$$I=\int \frac{1}{x^2 \sqrt{9x^2+4}}dx$$
$$\int \frac{\frac 23 \cosh(t)}{\frac 49\sinh^2(t) \sqrt{4\sinh^2(t)+4}}dt$$
$$\int \frac{\cosh(t)}{\frac 23\sinh^2(t) 2\cosh(t)}dt$$
$$I=\frac 34\int \frac{dt}{\sinh^2(t) }=-\frac 34\coth(t)$$
Substitute back now...
