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