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

Theorem

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

Proof
Let:

Also see

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