$\displaystyle \int 3xe^{2x^2} \ \mathrm{d}x$

How do you integrate this? Thanks in advance.

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- Jun 15th 2008, 03:38 AMSimplicityIntegration By Parts?
$\displaystyle \int 3xe^{2x^2} \ \mathrm{d}x$

How do you integrate this? Thanks in advance. - Jun 15th 2008, 03:41 AMMoo
Hello,

By the substitution : $\displaystyle t=2x^2$

:)

How can you know it's this way ? Because you can notice that the derivative of $\displaystyle 2x^2$ is $\displaystyle 4x$.

In front of the exponential, we have $\displaystyle 3x=\frac 34 (4x)$

So you have something like $\displaystyle k \int u'(x)e^{u(x)} dx$, k is a constant (here $\displaystyle \frac 34$)

Here, two methods :

- you substitute v=u(x)

- you recognize the derivative of $\displaystyle e^{u(x)}$

(Wink)