Solution
= Solution
Put $u=x^2$, so $x\,dx=du/2$. Then <integration by substitution> gives
$$
\int_0^\infty(x+x^3)e^{-x^2}dx=\frac12\int_0^\infty(1+u)e^{-u}du=\frac12(1+1)=\boxed{1}.
$$
The last two integrals are elementary exponential moments, or values of the <Gamma function>.