Evaluating $\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$ using a trigonometric substitution I have this integral:
$$\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$$
I tried to solve it with a trigonometric substitituon but I can't get any result.
I would appreciate if somebody could help me.
 A: Hint:
Put $x = \frac{3}{2}\tan\theta \Rightarrow dx = \frac{3}{2} \sec^2 \theta \ d\theta$, we have
\begin{align}
\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx&=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\cdot\sec^2\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{\sin^3\theta}{\cos^3\theta}\cdot\cos\theta\,\,d\theta\quad\Rightarrow\quad\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\mbox{and}\,\,\sec\theta=\frac{1}{\cos\theta}\\
&=\frac{3}{16}\int\frac{\sin^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{1-\cos^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\quad\Rightarrow\quad\mbox{set}\,\,t=\cos\theta \Rightarrow dt=-\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{t^2-1}{t^2}\,dt
\end{align}
Can you take it from here?
A: Hint: 
Make the substitution $x = \frac{3}{2}\tan\theta \Rightarrow dx = \frac{3}{2} \sec^2 \theta \ \ d\theta$.
A: If Trigonometric substitution is not mandatory, write $$x^3=\frac{x(4x^2+9)-9x}4$$
$$\implies\frac{x^3}{(4x^2+9)^{\frac32}}=\frac14\cdot\frac x{\sqrt{4x^2+9}}-\frac94\cdot\frac x{(4x^2+9)^{\frac32}}$$
Now write $4x^2+9=v$ or $\sqrt{4x^2+9}=u\implies4x^2+9=u^2$
