# Integration By Parts?

• Jun 15th 2008, 03:38 AM
Simplicity
Integration By Parts?
$\int 3xe^{2x^2} \ \mathrm{d}x$

How do you integrate this? Thanks in advance.
• Jun 15th 2008, 03:41 AM
Moo
Hello,

Quote:

Originally Posted by Air
$\int 3xe^{2x^2} \ \mathrm{d}x$

How do you integrate this? Thanks in advance.

By the substitution : $t=2x^2$
:)

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

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

So you have something like $k \int u'(x)e^{u(x)} dx$, k is a constant (here $\frac 34$)
Here, two methods :
- you substitute v=u(x)
- you recognize the derivative of $e^{u(x)}$

(Wink)