Answer
$$\cosh ^{-1} t =\ln (t+\sqrt{t^{2}-1})$$
Work Step by Step
Let $ t=\cosh x$.
Since
\begin{align*}
\cosh^2 x-\sinh^2 x&=1\\
\sinh x&=\sqrt{\cosh^2 x-1}\\
&= \sqrt{t^2-1}
\end{align*}
and
\begin{align*}
\sinh x+\cosh x&=\frac{e^{x}-e^{-x}}{2}+\frac{e^{x}+e^{-x}}{2}\\
&=e^{x}
\end{align*}
Then
\begin{align*}
\cosh ^{-1} t&=x\\
&=\ln (\sinh x+\cosh x)\\
&=\ln (t+\sqrt{t^{2}-1})
\end{align*}
