Answer
$$\int xe^{-x^2}dx=-\frac{e^{-x^2}}{2}+C$$
Work Step by Step
$$A=\int xe^{-x^2}dx$$
Let $u=-x^2$
Then $du=(-x^2)'dx=-2xdx$. So $xdx=-\frac{1}{2}du$.
Substitute into $A$, we have $$A=\int e^u(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int e^udu$$ $$A=-\frac{1}{2}e^u+C$$ $$A=-\frac{e^{-x^2}}{2}+C$$
