# Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form

## Theorem

$\displaystyle \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$

## Proof

 $\displaystyle \int \frac {\d x} {x \sqrt {a^2 - x^2} }$ $=$ $\displaystyle -\frac 1 a \sech^{-1} {\frac x a} + C$ Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$: $\sech^{-1}$ form $\displaystyle$ $=$ $\displaystyle -\frac 1 a \map \ln {\frac {1 + \sqrt {1 - \paren {\frac x a}^2} } {\frac x a} } + C$ Definition 2 of Inverse Hyperbolic Secant $\displaystyle$ $=$ $\displaystyle -\frac 1 a \map \ln {\frac {a + a \sqrt {1 - \paren {\frac x a}^2} } x} + C$ multiplying top and bottom by $a$ $\displaystyle$ $=$ $\displaystyle -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - a^2 \paren {\frac x a}^2} } x} + C$ moving $a$ within the square root $\displaystyle$ $=$ $\displaystyle -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$ simplifying

$\blacksquare$