$ \int\ \frac {dx} {x^4 - c^4} \\ $
is equal to (from integral tables)
$ \frac {1} {4 c^3} \ln \frac {x-c} {x+c} - \frac {1} {2 c^3} \tan^{-1}\frac {x} {c}$
If I let $\frac{x}{c}=\tan u$, then $dx/c= \sec^2 u du$, and the integral becomes $ \frac {1} {c^3} \int\ \frac {\sec ^2u du} {\tan^4u -1} = \frac {1} {c^3} \int\ \frac {\sec ^2u du} {(\tan^2u +1)(\tan^2u -1)} $ Since $\tan^2u+1=\sec^2u$ the integral becomes $ \frac {1} {c^3} \int\ \frac {du} {(\tan^2u -1)} $
Am I on the right track? What should I do next?