Answer
$$\frac{{e - 1}}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x{e^{{x^2}}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}xdx = \frac{1}{{2x}}du \cr
& {\text{The new limits of integration are}} \cr
& x = 1 \to u = 1 \cr
& x = 0 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_0^1 {x{e^{{x^2}}}} dx = \int_0^1 {x{e^u}\frac{1}{{2x}}du} \cr
& = \frac{1}{2}\int_0^1 {{e^u}du} \cr
& {\text{Integrating by tables}} \cr
& = \frac{1}{2}\left[ {{e^u}} \right]_0^1 \cr
& = \frac{1}{2}\left( {{e^1} - {e^0}} \right) \cr
& = \frac{{e - 1}}{2} \cr} $$
