# 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:

 $\displaystyle z$ $=$ $\displaystyle x^2$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle 2 x$ Power Rule for Derivatives $\displaystyle \leadsto \ \$ $\displaystyle \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }$ $=$ $\displaystyle \int \frac {\d z} {2 z \sqrt z \sqrt {z - a^2} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }$
$\displaystyle \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$

Setting:

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

 $\displaystyle \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }$ $=$ $\displaystyle \frac {-\sqrt {z - a^2} } {2 \paren {\paren {\frac 3 2} - 1} \paren {-a^2} z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} \paren {-a^2} } \int \frac {\d z} {z^{\paren {\frac 3 2} - 1} \sqrt{z - a^2} } + C$ 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 $x$ back for $z$

$\blacksquare$