Primitive of x over x fourth minus a fourth

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Theorem

$\ds \int \frac {x \rd x} {x^4 - a^4} = \frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C$


Proof

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {x \rd x} {x^4 - a^4}\) \(=\) \(\ds \int \frac {\d z} {2 \paren {z^2 - \paren {a^2}^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac 1 {2 a^2} \ln \size {\frac {z - a^2} {z + a^2} } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C\) substituting back for $z$ and simplifying

$\blacksquare$


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