Evaluate $\int xe^{-0.5x^2}dx$

Question

I tried to evaluate this term:

$$\int xe^{-0.5x^2}dx$$

but I couldn't find any way other than Wolfram Alpha.

How can I find the solution by hand?

Answer 1

we get $$-e^{-x^2/2}+C$$ it is easy to be seen

Answer 2

Hint: what is the derivative of $e^{-0.5x^2}$?

Answer 3

Put $-0.5x^2 = t$

On differentiate,

$-0.5 \times 2x^{2-1} dx = dt$

$-xdx = dt$

$xdx = -dt$

Then we have,

$- \int e^t dt$

On integration,

$-e^t + c$

Put value of t,

$-e^{-0.5x^2} + c$

Answer 4

Using the chain rule by setting $u \mapsto \sqrt{u} =: x$, we have $$ \int x e^{-x^{2}/2} = \int \sqrt{u}e^{-u/2}\cdot D\sqrt{u} = \int \sqrt{u}e^{-u/2}\frac{1}{2}\frac{1}{\sqrt{u}} = \frac{1}{2}\int e^{-u/2} = -\int D e^{-u/2} = -e^{-u/2} + \text{constant} = -e^{-x^{2}/2} + \text{constant}. $$