Answer
$$\frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) - \frac{\pi }{{12\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt 2 } {\frac{x}{{3 + {x^4}}}} dx \cr
& or \cr
& = \int_1^{\sqrt 2 } {\frac{x}{{3 + {{\left( {{x^2}} \right)}^2}}}} dx \cr
& {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \sqrt 2 ,{\text{ }}u = 2 \cr
& {\text{if }}x = 1,{\text{ }}u = 1 \cr
& {\text{so}} \cr
& = \int_1^{\sqrt 2 } {\frac{x}{{3 + {{\left( {{x^2}} \right)}^2}}}} dx = \int_1^2 {\frac{{1/2}}{{3 + {u^2}}}} du \cr
& = \frac{1}{2}\int_1^2 {\frac{{du}}{{{{\left( {\sqrt 3 } \right)}^2} + {u^2}}}} \cr
& {\text{find the antiderivative}}{\text{, using }}\int {\frac{1}{{{a^2} + {u^2}}}du = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C} {\text{ }} \cr
& = \frac{1}{2}\left. {\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right)} \right)} \right|_1^2 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{{2\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) - {{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)} \right) \cr
& {\text{simplify}} \cr
& = \frac{1}{{2\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) - \frac{\pi }{6}} \right) \cr
& = \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) - \frac{\pi }{{12\sqrt 3 }} \cr} $$
