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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n} = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^{n - 1} }$

for $x^2 > a^2$.

Proof
Let:

From Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$:
 * $\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$

Let:

Then:

Also see

 * Primitive of $\dfrac 1 {x \left({a^2 - x^2}\right)^n}$