<s>[INST] 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?
 [/INST] 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 $p\geq 2$,
$$(2p-2)\int\frac{dt}{(t^2+1)^p}=\frac{t}{(t^2+1)^{p-1}}
+(2p-3)\int\frac{dt}{(t^2+1)^{p-1}}.$$
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