$$\cosh(\sinh^{-1}x) = \sqrt{x^{2}+1}$$
I used the fact that $\cosh(x) = \frac{1}{2}(e^{x}+e^{-x})$
and that $\sinh^{-1}(x) = \ln(x + \sqrt{x^{2}+1})$
Eventually I simplified to
$$\frac{x^{2}+x\sqrt{x^{2}+1}+1}{x+ \sqrt{x^{2}+1}}$$
I verified and so far I am correct, but I can't see how I can simplify further.
Note that the top is $\sqrt{x^2+1}\left(x+\sqrt{x^2+1}\right)$.
Remark: You may have worked too hard by first obtaining a formula for $\sinh^{-1}x$, though such a formula is in fact useful.
For we have $\cosh^2 t-\sin^2 t=1$ for all $t$. Let $t=\sinh^{-1}(x)$. We get $\cosh^2 t=1+x^2$. But $\cosh t$ is always positive, so $\cosh t=\sqrt{1+x^2}$.