Answer
$$2\ln \left( 5 \right) - 4 + 2{\tan ^{ - 1}}2$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\ln \left( {{x^2} + 1} \right)} dx \cr
& {\text{substitute }}u = \ln \left( {{x^2} + 1} \right),{\text{ }}du = \frac{{2x}}{{{x^2} + 1}}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_0^2 {\ln \left( {{x^2} + 1} \right)} dx = \left. {\left( {x\ln \left( {{x^2} + 1} \right)} \right)} \right|_0^2 - \int_0^2 {\left( {\frac{{2{x^2}}}{{{x^2} + 1}}} \right)dx} \cr
& {\text{long division}} \cr
& \int_0^2 {\ln \left( {{x^2} + 1} \right)} dx = \left. {\left( {x\ln \left( {{x^2} + 1} \right)} \right)} \right|_0^2 - \int_0^2 {\left( {2 - \frac{2}{{{x^2} + 1}}} \right)dx} \cr
& {\text{integrating}} \cr
& \int_0^2 {\ln \left( {{x^2} + 1} \right)} dx = \left. {\left( {x\ln \left( {{x^2} + 1} \right)} \right)} \right|_0^2 - \left. {\left( {2x - 2{{\tan }^{ - 1}}x} \right)} \right|_0^2 \cr
& \int_0^2 {\ln \left( {{x^2} + 1} \right)} dx = \left. {\left( {x\ln \left( {{x^2} + 1} \right) - 2x + 2{{\tan }^{ - 1}}x} \right)} \right|_0^2 \cr
& {\text{evaluate limits}} \cr
& = \left( {2\ln \left( {{2^2} + 1} \right) - 2\left( 2 \right) + 2{{\tan }^{ - 1}}2} \right) - \left( {2\ln \left( {{0^2} + 1} \right) - 2\left( 0 \right) + 2{{\tan }^{ - 1}}0} \right) \cr
& {\text{simplify}} \cr
& = \left( {2\ln \left( 5 \right) - 4 + 2{{\tan }^{ - 1}}2} \right) - \left( 0 \right) \cr
& = 2\ln \left( 5 \right) - 4 + 2{\tan ^{ - 1}}2 \cr} $$
