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

## Theorem

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

## Proof

Let:

 $\displaystyle z$ $=$ $\displaystyle x^2$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d z} {\mathrm d x}$ $=$ $\displaystyle 2 x$ Power Rule for Derivatives $\displaystyle \implies \ \$ $\displaystyle \int \frac {\mathrm d x} {x^2 \sqrt {x^2 - a^2} }$ $=$ $\displaystyle \int \frac {\mathrm d z} {2 z \sqrt z \sqrt {z - a^2} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z - a^2} }$
$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$

Setting:

 $\displaystyle x$ $:=$ $\displaystyle z$ $\displaystyle m$ $:=$ $\displaystyle \frac 3 2$ $\displaystyle a$ $:=$ $\displaystyle 1$ $\displaystyle b$ $:=$ $\displaystyle -a^2$

 $\displaystyle \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z - a^2} }$ $=$ $\displaystyle \frac {-\sqrt{z - a^2} } {2 \left({\left({\frac 3 2}\right) - 1}\right) \left({-a^2}\right) z^{\left({3/2}\right) - 1} } - \frac {2 \left({\frac 3 2}\right) - 3} {2 \left({2 \left({\frac 3 2}\right) - 2}\right) \left({-a^2}\right)} \int \frac {\mathrm d z} {z^{\left({\frac 3 2}\right) - 1} \sqrt{z - a^2} } + C$ Primitive of $\dfrac 1{x^m \sqrt{a x + b} }$ $\displaystyle$ $=$ $\displaystyle \frac {-\sqrt{z - a^2} } {-a^2 z^{1/2} } - 0 + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$ substituting back for $z$

$\blacksquare$