# Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form

## Theorem

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

### Corollary

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

## Proof 1

 $\ds \int \frac {\d x} {\sqrt {x^2 + a^2} }$ $=$ $\ds \arsinh {\frac x a} + C$ Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ in $\arsinh$ form $\ds$ $=$ $\ds \map \ln {x + \sqrt {x^2 + a^2} } - \ln a + C$ $\arsinh \dfrac x a$ in Logarithm Form $\ds$ $=$ $\ds \map \ln {x + \sqrt {x^2 + a^2} } + C$ subsuming $-\ln a$ into arbitrary constant

$\blacksquare$

## Proof 2

Let $y^2 = a^2 + x^2$.

Then:

$\ds 2 y \frac {\d y} {\d x}$ $=$ $\ds 2 x$ Power Rule for Derivatives, Chain Rule for Derivatives
$\ds \leadsto \ \$ $\ds y \frac {\d y} {\d x}$ $=$ $\ds x$ simplification
$\ds \leadsto \ \$ $\ds \frac {\d y} x$ $=$ $\ds \frac {\d x} y$
$\ds$ $=$ $\ds \frac {\d x + \d y} {x + y}$

$\ds \leadsto \ \$ $\ds \int \frac {\d x} {\sqrt {x^2 + a^2} }$ $=$ $\ds \int \frac {\d x} y$ substituting for $y$
$\ds$ $=$ $\ds \frac {\d x + \d y} {x + y}$ from above
$\ds$ $=$ $\ds \ln \size {x + y} + C$ Primitive of Function under its Derivative
$\ds$ $=$ $\ds \ln \size {x + \sqrt {x^2 + a^2} } + C$ substituting back
$\ds$ $=$ $\ds \map \ln {x + \sqrt {x^2 + a^2} } + C$ argument of $\ln$ always positive

$\blacksquare$

## Also presented as

Some sources present this in the form:

$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {\dfrac {x + \sqrt {x^2 + a^2} } a} + C$

which is the same as above, except that the constant $a$ has not been subsumed into the arbitrary constant $C$.