Answer
$y=ln(x\sqrt {1+x^2})$
Work Step by Step
Let suppose $y=sinh^{-1}x$
$x=sinh$ $y$
The example 1(a) shows that
$cosh^2$ $ y-sinh^2 $ $ y=1$
As we know from exercise $9$.
$e^y=cosh$ $ y+sinh $ $ y$
$=x+\sqrt {1+x^2}$
$y=ln(x\sqrt {1+x^2})$