How to show that $\int x e^{-x^2/2} dx = -e^{-x^2/2} + c$?

Question

I wish to show that

$\int x e^{-x^2/2} dx = -e^{-x^2/2} + c$

I tried using integration by parts

Consider $\int u dv = vu - \int v du$

Let $dv = xdx, u = e^{-x^2/2}$

Then $\int x e^{-x^2/2} dx = x^2/2 * e^{-x^2/2} + \int x^2/2 * xe^{-x^2/2} dx$

This gets me no where close to my desired solution

Can anyone help?

Answer 1

Go by substitution. Put $\dfrac{x^2}{2} = t$

$$\therefore x dx = dt$$

$$\int x e^{-\frac{x^2}{2}} dx = \int e^{-t} dt = -e^{-t} + c$$

On resubstituting, we get $\int x e^{-\frac{x^2}{2}} dx = -e^{-\frac{x^2}{2}} + c$

Answer 2

Don't use integration by parts. This is just a u-substitution. Sub $u = x^2/2$ so that $du = xdx$ is right there.