Can't compute this integral I would appreciate your help solving this integral:
$$\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x$$
I tried using linear substitution with $ t = x/a $ and then trying to bring it to some combination of the known integral of $\arctan (x) = \frac{1}{x^2+1}$ but I'm not sure it will be helpful because there isn't just $1 $ in the numerator.
Basically, I got stuck very early in the process:
$$\int_2^3 \frac{1+x^3}{(a^2(\frac{x^2}{a^2}+1))^\frac{3}{2}}\mathrm{d}x$$
Thank you.
 A: I suggest breaking it into two integrals, and using the trig substitution $x=a\tan t$ to turn each one into a trigonometric integral. Thus:
$$\begin{align}
\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x &= \int_2^3 \frac{dx}{(x^2+a^2)^{3/2}} + \int_2^3\frac{x^3}{(x^2+a^2)^{3/2}}dx\\
&=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sec^2 t}{a^3\sec^3 t}dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a^3 \tan^3 t \cdot a\sec^2 t}{a^3\sec^3 t}dt\\
&=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{1}{a^2}\cos t dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sin^3 t}{\cos^2 t}dt
\end{align}$$
Can you get it from there?
A: Hint: Let $x \mapsto a \tan \theta$, then $\mathrm{d}x = a(1 + \tan \theta) \,\mathrm{d} \theta$. Then
$$
\int \frac{1+x^3}{(a^2+x^2)^{3/2}} = \frac{1}{a^2}\int \frac{\sin^3\theta}{\cos ^2\theta}+\cos \theta\,\mathrm{d}\theta 
$$
For the first integral use $\sin^3 \theta = \sin \theta(1-\cos^2\theta)$ and set $y \mapsto \cos \theta$. The latter integral is trivial. 
A: Your integral can be split into $$\int\frac{dx}{(x^2+a^2)^{3/2}}+\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}}$$
The first integral can be solved by trig substitution. U-substitution can be used on the second integral to obtain
$$\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}}
=\frac{1}{2}\int\frac{u-a^2}{u^{3/2}}du \\ 
= \frac{1}{2}\int(u^{-1/2}-a^2u^{-3/2})du$$.
And it's busywork from there.
