# Primitive of Root of x squared plus a squared over x squared

## Theorem

$\displaystyle \int \frac {\sqrt {x^2 + a^2} } {x^2} \ \mathrm d x = \frac {-\sqrt {x^2 + a^2} } x + \ln \left({x + \sqrt {x^2 + a^2} }\right) + C$

## 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 \ \$ $\displaystyle \int \frac {\sqrt {x^2 + a^2} } {x^2} \ \mathrm d x$ $=$ $\displaystyle \int \frac {\sqrt {z + a^2} \ \mathrm d z} {2 z \sqrt z}$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 2 \int \frac {\sqrt {z + a^2} \ \mathrm d z} {z^{3/2} }$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac 1 2 \left({\frac {-\sqrt {z + a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\mathrm d z} {\sqrt z \sqrt {z + a^2} } }\right) + C$ Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$ $\displaystyle$ $=$ $\displaystyle \frac {-\sqrt {x^2 + a^2} } x + \frac 1 2 \int \frac {2 x \ \mathrm d x} {x \sqrt {x^2 + a^2} } + C$ substituting for $z$ $\displaystyle$ $=$ $\displaystyle \frac {-\sqrt {x^2 + a^2} } x + \int \frac {\mathrm d x} {\sqrt {x^2 + a^2} } + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {-\sqrt {x^2 + a^2} } x + \ln \left({x + \sqrt {x^2 + a^2} }\right) + C$ Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$

$\blacksquare$