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