Primitive of Reciprocal of x by Root of x squared minus a squared cubed

Theorem

 * $\displaystyle \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C$

for $\size x > a$.

Proof
Let:

Also see

 * Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 + a^2} }^3}$
 * Primitive of $\dfrac 1 {x \paren {\sqrt {a^2 - x^2} }^3}$