Answer
$$x\ln \left( {{x^2} + 4} \right) - 2x + 4{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \left( {{x^2} + 4} \right)} dx \cr
& {\text{substitute }}u = \ln \left( {{x^2} + 4} \right),{\text{ }}du = \frac{{2x}}{{{x^2} + 4}}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - \int {x\left( {\frac{{2x}}{{{x^2} + 4}}} \right)dx} \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - \int {\frac{{2{x^2}}}{{{x^2} + 4}}dx} \cr
& {\text{long division}} \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - \int {\left( {2 - \frac{8}{{{x^2} + 4}}} \right)dx} \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - \int {2dx} + \int {\frac{8}{{{x^2} + 4}}} \cr
& {\text{find antiderivatives}} \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - 2x + \frac{8}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& \int {\ln \left( {{x^2} + 4} \right)} dx = x\ln \left( {{x^2} + 4} \right) - 2x + 4{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
