Primitive of x squared over x fourth plus a fourth

Theorem

 * $\displaystyle \int \frac {x^2 \rd x} {x^4 + a^4} = \frac 1 {4 a \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } - \frac 1 {2 a \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }$

Proof
Then:

Similarly:

Thus: