Answer
$$\frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{x}{{{x^4} + 1}}dx} \cr
& {\text{Rewrite}} \cr
& \int_0^1 {\frac{x}{{{x^4} + 1}}dx} = \int_0^1 {\frac{x}{{{{\left( {{x^2}} \right)}^2} + 1}}dx} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& {\text{The new limits are:}} \cr
& x = 0 \Rightarrow u = 0 \cr
& x = 1 \Rightarrow u = 1 \cr
& {\text{,then}} \cr
& \int_0^1 {\frac{x}{{{{\left( {{x^2}} \right)}^2} + 1}}dx} = \frac{1}{2}\int_0^1 {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left[ {{{\tan }^{ - 1}}u} \right]_0^1 \cr
& = \frac{1}{2}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right] \cr
& = \frac{1}{2}\left( {\frac{\pi }{4}} \right) \cr
& = \frac{\pi }{8} \cr} $$
