# Primitive of Inverse Hyperbolic Sine of x over a

## Theorem

$\displaystyle \int \sinh^{-1} \frac x a \rd x = x \sinh^{-1} \frac x a - \sqrt {x^2 + a^2} + C$

## Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\displaystyle u$ $=$ $\displaystyle \sinh^{-1} \frac x a$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac 1 {\sqrt {x^2 + a^2} }$ Derivative of $\sinh^{-1} \dfrac x a$

and let:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle v$ $=$ $\displaystyle x$ Primitive of Constant

Then:

 $\displaystyle \int \sinh^{-1} \frac x a \rd x$ $=$ $\displaystyle x \sinh^{-1} \frac x a - \int x \paren {\frac 1 {\sqrt {x^2 + a^2} } } \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle x \sinh^{-1} \frac x a - \int \frac {x \rd x} {\sqrt {x^2 + a^2} } + C$ simplifying $\displaystyle$ $=$ $\displaystyle x \sinh^{-1} \frac x a - \sqrt {x^2 + a^2} + C$ Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$

$\blacksquare$