Answer
$$x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \sqrt {{x^2} - 1} + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = \ln \left( {x + \sqrt {{x^2} - 1} } \right),{\text{ }} \cr
& du = \frac{{1 + \frac{{2x}}{{2\sqrt {{x^2} - 1} }}}}{{x + \sqrt {{x^2} - 1} }}dx \cr
& {\text{Simplifying}} \cr
& du = \frac{{2\sqrt {{x^2} - 1} + 2x}}{{2\sqrt {{x^2} - 1} \left( {x + \sqrt {{x^2} - 1} } \right)}}dx \cr
& du = \frac{{2\left( {\sqrt {{x^2} - 1} + x} \right)}}{{2\sqrt {{x^2} - 1} \left( {x + \sqrt {{x^2} - 1} } \right)}}dx \cr
& du = \frac{1}{{\sqrt {{x^2} - 1} }}dx \cr
& dv = dx \to v = x \cr
& {\text{Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \int {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr
& {\text{Integrating}} \cr
& \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \sqrt {{x^2} - 1} + C \cr} $$
