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?

$\displaystyle\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$
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.
