Answer
$${\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {\sinh ^{ - 1}}x,{\text{ then}} \cr
& y = {\sinh ^{ - 1}}x \Leftrightarrow x = \sinh y \cr
& {\text{Using the hyperbolic identities}} \cr
& x = \sinh y = \frac{{{e^y} - {e^{ - y}}}}{2} \cr
& x = \frac{{{e^y} - {e^{ - y}}}}{2} \cr
& 2x = {e^y} - {e^{ - y}} \cr
& {e^y} - 2x - {e^{ - y}} = 0 \cr
& {\text{Multiply the equation by }}{e^y} \cr
& {e^{2y}} - 2x{e^y} - 1 = 0 \cr
& {\text{By the quadratic formula}} \cr
& {e^y} = \frac{{ - \left( { - 2x} \right) \pm \sqrt {{{\left( {2x} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{2} \cr
& {e^y} = \frac{{2x \pm \sqrt {4{x^2} + 4} }}{2} \cr
& {e^y} = \frac{{2x \pm 2\sqrt {{x^2} + 1} }}{2} \cr
& {e^y} = x \pm \sqrt {{x^2} + 1} \cr
& {\text{Taking the positive square root because }}{e^y} > 0{\text{ for all real }}x \cr
& {e^y} = x + \sqrt {{x^2} + 1} \cr
& {\text{Solve for }}y{\text{ by taking the natural logarithm of both sides}} \cr
& y = \ln \left( {x + \sqrt {{x^2} + 1} } \right) \cr
& {\text{As we know }}y = {\sinh ^{ - 1}}x,{\text{ then}} \cr
& {\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right) \cr} $$
