<s>[INST] Steps in evaluating $\int \frac{1}{x^4-c^4} dx$? $$ \int\ \frac {dx} {x^4 - c^4} \\ $$
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?
 [/INST] You’d be better off writing factoring $x^4-c^4$ as $(x^2-c^2)(x^2+c^2)$ and then as $(x-c)(x+c)(x^2+c^2)$ and decomposing $\frac1{x^4-c^4}$ into partial fractions:
$$\frac1{x^4-c^4}=\frac{A}{x-c}+\frac{B}{x+c}+\frac{Cx+D}{x^2+c^2}\;.$$
Once you’ve done the algebra to find $A,B$, and $C$, the integrations should be pretty straightforward: two logs and a tangent substitution.
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