Primitive of Reciprocal of x by x fourth plus a fourth

Theorem

$\ds \int \frac {\d x} {x \paren {x^4 + a^4} } = \frac 1 {4 a^4} \map \ln {\frac {x^4} {x^4 + a^4} }$

$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$

So:

$\ds \int \frac {\d x} {x \paren {x^4 + a^4} }$

$=$

$\ds \frac 1 {4 a^4} \ln \size {\frac {x^4} {x^4 + a^4} } + C$

$\ds $

$=$

$\ds \frac 1 {4 a^4} \map \ln {\frac {x^4} {x^4 + a^4} } + C$

directly.

$\blacksquare$