Answer
$$\frac{\pi }{8} - \frac{1}{4}\ln 2$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {{{\tan }^{ - 1}}\left( {2x} \right)} dx \cr
& {\text{substitute }}u = {\tan ^{ - 1}}\left( {2x} \right),{\text{ }}du = \frac{3}{{1 + {{\left( {2x} \right)}^2}}}dx \cr
& du = \frac{2}{{1 + 4{x^2}}}dx \cr
& dv = dx, \cr
& v = x \cr
& {\text{using integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{we have}} \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \int {\left( x \right)\left( {\frac{2}{{1 + 4{x^2}}}} \right)dx} \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \int {\frac{{2x}}{{1 + 4{x^2}}}dx} \cr
& {\text{find antiderivative}}{\text{, }}t = 1 + 4{x^2}{\text{ }}dt = 8xdx \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \frac{2}{8}\int {\frac{{dt}}{t}} \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\int {\frac{{dt}}{t}} \cr
& {\text{replace }}t = 1 + 4{x^2} \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\ln \left| {1 + 4{x^2}} \right| + C \cr
& \int {{{\tan }^{ - 1}}\left( {2x} \right)} dx = x{\tan ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\ln \left( {1 + 4{x^2}} \right) + C \cr
& \int_0^{1/2} {{{\tan }^{ - 1}}\left( {2x} \right)} dx = \left[ {x{{\tan }^{ - 1}}\left( {2x} \right) - \frac{1}{4}\ln \left( {1 + 4{x^2}} \right)} \right]_0^{1/2} \cr
& {\text{fundamental theorem}} \cr
& = \left[ {\frac{1}{2}{{\tan }^{ - 1}}\left( {2\left( {\frac{1}{2}} \right)} \right) - \frac{1}{4}\ln \left( {1 + 4{{\left( {\frac{1}{2}} \right)}^2}} \right)} \right] - \left[ 0 \right] \cr
& = \left[ {\frac{1}{2}{{\tan }^{ - 1}}\left( 1 \right) - \frac{1}{4}\ln \left( {1 + 1} \right)} \right] \cr
& = \frac{\pi }{8} - \frac{1}{4}\ln 2 \cr} $$
