Primitive of Reciprocal of x by Power of x squared plus a squared

Theorem

 * $\ds \int \frac {\d x} {x \paren {x^2 + a^2}^n} = \frac 1 {2 \paren {n - 1} a^2 \paren {x^2 + a^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {x^2 + a^2}^{n - 1} }$

Proof
Let:

From Reduction Formula for Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
 * $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$

Let:

Then: