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$$ \int\ \frac {dx} {x^4 - c^4} \\ $$

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?

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migrated from Feb 8 '12 at 10:33

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Why not just use partial faction decomposition? – anon Feb 8 '12 at 10:41
Thank you very much for your hint!! – Tony Feb 8 '12 at 11:06
up vote 8 down vote accepted

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:


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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Thank you very much!! Very grateful for your detailed response!! – Tony Feb 8 '12 at 11:05
Tony: The partial fractions is really easy. Don't do $A$, $B$, $C$, $D$. Note first that your function is, apart from a constant factor, $\frac{1}{x^2-c^2}-\frac{1}{x^2+c^2}$. The first part has basically the same simple decomposition, and we are finished. – André Nicolas Feb 8 '12 at 11:11
Your hint simplifies the problem a lot!! Thank you very much for your valuable hint!! – Tony Feb 8 '12 at 11:16
There is a hazard of looking up the "answer" first, then trying to get it. – GEdgar Feb 8 '12 at 15:36

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