Answer
$$\ln \left( 5 \right) - 2 + 4{\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\ln \left( {4 + {x^2}} \right)dx} \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = \ln \left( {4 + {x^2}} \right),{\text{ }}du = \frac{{2x}}{{4 + {x^2}}}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{ Integration by Parts Formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\ln \left( {4 + {x^2}} \right)} dx = x\ln \left( {4 + {x^2}} \right) - \int {\frac{{2{x^2}}}{{4 + {x^2}}}dx} \cr
& {\text{By long division }}\frac{{2{x^2}}}{{4 + {x^2}}} = 2 - \frac{8}{{4 + {x^2}}} \cr
& \int {\ln \left( {4 + {x^2}} \right)} dx = x\ln \left( {4 + {x^2}} \right) - \int {\left( {2 - \frac{8}{{4 + {x^2}}}} \right)} dx \cr
& \int {\ln \left( {4 + {x^2}} \right)} dx = x\ln \left( {4 + {x^2}} \right) - 2\int {dx} + 8\int {\frac{1}{{4 + {x^2}}}dx} \cr
& {\text{Integrate}} \cr
& \int {\ln \left( {4 + {x^2}} \right)} dx = x\ln \left( {4 + {x^2}} \right) - 2x + 4{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\ln \left( {4 + {x^2}} \right)} dx = \left[ {x\ln \left( {4 + {x^2}} \right) - 2x + 4{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^1 \cr
& = \left[ {\ln \left( 5 \right) - 2 + 4{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right)} \right] - \left[ {0\ln \left( 4 \right) - 0 + 4{{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& {\text{Simplify}} \cr
& = \ln \left( 5 \right) - 2 + 4{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) \cr} $$
