Primitive of Reciprocal of x by x fourth minus a fourth

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Theorem

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


Proof

From Primitive of $\dfrac 1 {x \paren {x^n - a^n} }$:

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


Setting $n = 4$:

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

directly.

$\blacksquare$


Sources