# Attempt to solve confusing integral $x^3 *e^[x^2] * dx (See attached scanned paper) Printable View • Jan 15th 2011, 07:14 PM Riazy Attempt to solve confusing integral$x^3 *e^[x^2] * dx (See attached scanned paper)
Here it is Attachment 20455
• Jan 15th 2011, 07:22 PM
tonio
Quote:

Originally Posted by Riazy
Here it is Attachment 20455

Put $\displaystyle{u = x^2\,; \;u'=2x\,,\,\,v'=xe^{x^2}\,;\;v=\frac{1}{2}e^{x^2} }$ , so integrating by parts:

$\displaystyle{\int x^3e^{x^2} dx=\int x^2(xe^{x^2})dx=\frac{1}{2}x^2e^{x^2}-\int xe^{x^2}dx=\frac{e^{x^2}}{2}\left(x^2-1)+C$ .

Tonio
• Jan 15th 2011, 07:25 PM
Prove It
$\displaystyle \int{x^3e^{x^2}\,dx} = \int{\frac{1}{2}x^2\cdot 2x\,e^{x^2}\,dx}$.

Now use integration by parts with $\displaystyle u = \frac{1}{2}x^2 \implies du = x\,dx$ and $\displaystyle dv = 2x\,e^{x^2}\,dx \implies v = e^{x^2}$ and the integral becomes

$\displaystyle \frac{1}{2}x^2e^{x^2} - \int{x\,e^{x^2}\,dx}$

$\displaystyle = \frac{1}{2}x^2e^{x^2} - \frac{1}{2}\int{2x\,e^{x^2}\,dx}$

$\displaystyle = \frac{1}{2}x^2e^{x^2} - \frac{1}{2}e^{x^2} + C$.