Solving Integration Problem: Finding Int. of x^2e^(-x^2)

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I feel really dumb for asking this, because I know it's something simple I'm just not seeing. Ok, given that

\int _{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi }

how to I find

\int _{-\infty}^{\infty} x^2e^{-x^2}dx = ?

I have tried the substitution u=x^2, and integration by parts, but nothing is working. Any help? Thanks
 
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The easy way to do this problem is to generalize your first integral. Can you show the integral of exp(-ax^2) is sqrt(pi/a) for a>0? (Use a substitution x=sqrt(a)*u). Now differentiate that with respect to a. Finally put a=1 again.
 
This might help:

d/dx (x*[e^-x^2]) = ...
Solve it and then integrate!
 
Integration by parts should work. You have

<br /> \int_{-\infty}^\infty x^2 e^{-x^2} \, dx = \sqrt \pi<br />

Set

<br /> u = x, \quad dv =x e^{-x^2} dx<br />

Then

<br /> \int u \, dv = uv - \int v \, du<br />

should, with careful attention to the uv term at the infinities, work fine.
 
Thanks everyone.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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