Inverting Cosh x
\[cosh^{-1}x\]
let \[y=coshx=\frac{e^x+e^{-x}}{2}\]
the multiplying by \[2e^x\]
gives\[2e^xy=e^{2x}+1 \rightarrow 2y(e^x)-e^{2x}-1=0 \rightarrow e^{2x}-2ye^x+1=0\]
This is a quadratic equation in
\[e^x\]
.\[e^x= \frac{2y \pm \sqrt{(-2y)^2- 4 \times 1 \times 1}}{2} = y \pm \sqrt{y^2-1}\]
.Hence
\[x= ln(y \pm \sqrt{y^2-1})= ln(y \pm \sqrt{y^2+1})\]
.Only the positive square root will return a valid value since
\[x, \; coshx\]
, increase together which is impossible with the negative square root so \[x= ln(y + \sqrt{y^2-1})\]
.Then
\[cosh^{-1}x= ln(x + \sqrt{x^2-1})\]
. 
