Answer
$$\frac{1}{2}\left( {2 + {x^2}} \right)\left[ {\ln \left( {2 + {x^2}} \right) - 1} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {x\ln \left( {2 + {x^2}} \right)} dx \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = 2 + {x^2},\,\,\,\,\,\,\,du = 2xdx,\,\,\,\,\,\,dx = \frac{{du}}{{2x}} \cr
& {\text{write in terms of }}u \cr
& \int {x\ln \left( {2 + {x^2}} \right)} dx = \int {x\ln u} \left( {\frac{{du}}{{2x}}} \right) \cr
& = \frac{1}{2}\int {\ln u} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 11}} \cr
& \left( {11} \right):\,\,\,\,\,\,\int {\ln u} du = u\ln u - u + C \cr
& \frac{1}{2}\int {\ln u} du = \frac{1}{2}u\ln u - \frac{1}{2}u + C \cr
& {\text{write in terms of }}x{\text{; replace }}2 + {x^2}{\text{ for }}u \cr
& = \frac{1}{2}\left( {2 + {x^2}} \right)\ln \left( {2 + {x^2}} \right) - \frac{1}{2}\left( {2 + {x^2}} \right) + C \cr
& {\text{factoring}} \cr
& = \frac{1}{2}\left( {2 + {x^2}} \right)\left[ {\ln \left( {2 + {x^2}} \right) - 1} \right] + C \cr} $$