Primitive of x over Power of a squared minus x squared

Theorem

 * $\displaystyle \int \frac {x \rd x} {\paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} }$

for $x^2 < a^2$.

Proof
Let:

Also see

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