Primitive of Reciprocal of Power of a squared minus x squared

Theorem

 * $\ds \int \frac {\d x} {\paren {a^2 - x^2}^n} = \frac x {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }$

for $x^2 > a^2$.

Proof
Aiming for an expression in the form:
 * $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$

in order to use the technique of Integration by Parts, let:

Thus:

Then:

Also see

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