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

Jump to navigation Jump to search

## Theorem

$\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^2} \rd x = \frac {-\sqrt {x^2 - a^2} } x + \ln \size {x + \sqrt {x^2 - a^2} } + C$

for $\size x \ge a$.

## Proof

Let:

 $\ds z$ $=$ $\ds x^2$ $\ds \leadsto \ \$ $\ds \frac {\d z} {\d x}$ $=$ $\ds 2 x$ Power Rule for Derivatives $\ds \leadsto \ \$ $\ds \int \frac {\sqrt {x^2 - a^2} } {x^2} \rd x$ $=$ $\ds \int \frac {\sqrt {z - a^2} \rd z} {2 z \sqrt z}$ Integration by Substitution $\ds$ $=$ $\ds \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} {z^{3/2} }$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac 1 2 \paren {\frac {-\sqrt {z - a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {z - a^2} } } + C$ Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$ $\ds$ $=$ $\ds \frac {-\sqrt {x^2 - a^2} } x + \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {x^2 - a^2} } + C$ substituting for $z$ $\ds$ $=$ $\ds \frac {-\sqrt {x^2 - a^2} } x + \int \frac {\d x} {\sqrt {x^2 - a^2} } + C$ simplifying $\ds$ $=$ $\ds \frac {-\sqrt {x^2 - a^2} } x + \ln \size {x + \sqrt {x^2 - a^2} } + C$ Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$

$\blacksquare$