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

Theorem

 * $\displaystyle \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$

for $\size x > a$.

Proof
Let:

Using Primitive of $\dfrac 1 {u^m \sqrt {a u + b} }$:


 * $\displaystyle \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$

Setting:

Also see

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