Answer
$$\frac{1}{4}{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) - \frac{1}{4}\ln \,\left( {\frac{5}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy \cr
& {\text{substitute }}u = {\tan ^{ - 1}}{y^2},{\text{ }}du = \frac{{2y}}{{1 + {{\left( {{y^2}} \right)}^2}}}dy,{\text{ }}du = \frac{{2y}}{{1 + {y^4}}}dy \cr
& dv = ydy,{\text{ }}v = \frac{{{y^2}}}{2} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy = \left. {\left( {\frac{{{y^2}}}{2}{{\tan }^{ - 1}}{y^2}} \right)} \right|_0^{1/\sqrt 2 } - \int_0^{1/\sqrt 2 } {\left( {\frac{{{y^2}}}{2}} \right)\left( {\frac{{2y}}{{1 + {y^4}}}} \right)dy} \cr
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy = \left. {\left( {\frac{{{y^2}}}{2}{{\tan }^{ - 1}}{y^2}} \right)} \right|_0^{1/\sqrt 2 } - \int_0^{1/\sqrt 2 } {\left( {\frac{{{y^3}}}{{1 + {y^4}}}} \right)dy} \cr
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy = \left. {\left( {\frac{{{y^2}}}{2}{{\tan }^{ - 1}}{y^2}} \right)} \right|_0^{1/\sqrt 2 } - \frac{1}{4}\int_0^{1/\sqrt 2 } {\left( {\frac{{4{y^3}}}{{1 + {y^4}}}} \right)dy} \cr
& {\text{integrating}} \cr
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy = \left. {\left( {\frac{{{y^2}}}{2}{{\tan }^{ - 1}}{y^2}} \right)} \right|_0^{1/\sqrt 2 } - \frac{1}{4}\left. {\left( {\ln \,\left( {1 + {y^4}} \right)} \right)} \right|_0^{1/\sqrt 2 } \cr
& \int_0^{1/\sqrt 2 } {y{{\tan }^{ - 1}}{y^2}} dy = \left. {\left( {\frac{{{y^2}}}{2}{{\tan }^{ - 1}}{y^2} - \frac{1}{4}\ln \,\left( {1 + {y^4}} \right)} \right)} \right|_0^{1/\sqrt 2 } \cr
& {\text{evaluate limits}} \cr
& = \left[ {\frac{{{{\left( {1/\sqrt 2 } \right)}^2}}}{2}{{\tan }^{ - 1}}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} - \frac{1}{4}\ln \,\left( {1 + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} \right)} \right] - \left[ 0 \right] \cr
& {\text{Simplify}} \cr
& = \left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{1}{2}} \right) - \frac{1}{4}\ln \,\left( {\frac{5}{4}} \right)} \right] - \left[ 0 \right] \cr
& = \frac{1}{4}{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) - \frac{1}{4}\ln \,\left( {\frac{5}{4}} \right) \cr} $$
