Definite integral for a 4 degree function 
The integral is:
  $$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$

I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
 A: Let $x=at$ and by Partial-Fraction Decomposition we get 
\begin{align*}\int_0^a \frac{x^4}{(x^2+a^2)^4}dx&=\frac{1}{a^3}\int_0^1 \frac{t^4}{(t^2+1)^4}dt\\
&=\frac{1}{a^3}\int_0^1\left(\frac{1}{(t^2+1)^2}-\frac{2}{(t^2+1)^3}+\frac{1}{(t^2+1)^4}\right)\,dt
\\&=\frac{1}{16a^3}\left[\arctan(t)
+\frac{t}{t^2+1}-\frac{(14/3)t}{(t^2+1)^2}+\frac{(8/3)t}{(t^2+1)^3}\right]_0^1\\&=\frac{3\pi-4}{192a^3}.
\end{align*}
P.S. Note that for $p\geq 2$,
$$(2p-2)\int\frac{dt}{(t^2+1)^p}=\frac{t}{(t^2+1)^{p-1}}
+(2p-3)\int\frac{dt}{(t^2+1)^{p-1}}.$$
A: One way to solve this problem is to do integration by parts.
$$\int{x^3 \frac{x}{(x^2+a^2)^4}dx}=-\frac{x^3}{6(x^2+a^2)^3}+\frac{1}{2}\int{x\frac{x}{(x^2+a^2)^3} dx}.$$
Continuing further
$$\int{x\frac{x}{(x^2+a^2)^3} dx} = -\frac{1}{4} \frac{x}{(x^2+a^2)^2}+\frac{1}{4}\int{\frac{1}{(x^2+a^2)^2}dx}.$$
Now you can substitute $x=a\tan{\theta}$ to get the final result.
A: Hint
You obtain the integral: $\;\displaystyle\int_0^\tfrac\pi4\!\frac{\tan^4\theta}{a^3(1+\tan^2\theta)^3}\,\mathrm d\mkern1mu\theta=\frac1{a^3}\int_0^\tfrac\pi4\frac{\sin^4\theta}{\cos\theta}\,\mathrm d\mkern1mu\theta$.
Bioche's rules say you should set $u=\sin\theta$.
A: $\displaystyle\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$
Where do we get with the substitution you have suggested?
$x = a\tan\theta\\
dx = a\sec^2\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{(a^4\tan^4\theta)(a\sec^2\theta)}{(a^2\tan^2\theta+a^2)^4}d\theta\\
$
Looks promising:
Keep simplifying
$\displaystyle\int_0^{\frac \pi 4} \frac{a^5\tan^4\theta\sec^2\theta}{a^8\sec^8\theta}d\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{\tan^4\theta}{a^3\sec^6\theta}d\theta$
Let's state this into terms of $\sin\theta, \cos\theta$
$\displaystyle\frac 1{a^3}\int_0^{\frac \pi 4} \sin^4\theta\cos^2\theta\ d\theta$
You need to apply your half angle identities, perhaps repeatedly.
$\displaystyle\sin^2\theta = \frac 12 (1-\cos 2\theta), \cos^2\theta = \frac 12 (1+\cos\theta) $
$\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} (1-\cos 2\theta)^2(1+\cos 2\theta)\ d\theta\\
\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} [1-\cos 2\theta -\cos^2 2\theta + \cos^3 2\theta] \ d\theta\\
\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} [1-\cos 2\theta -\frac 12 (1+\cos 4\theta) + \cos 2\theta (1-\sin^2 2\theta)]\ d\theta\\$
And that looks pretty straight-forward.
