# integration by parts

• Jun 13th 2007, 10:38 AM
viet
integration by parts
1) $\displaystyle \int xe^{2x}dx$

$\displaystyle u = x, dv = e^{2x}dx$
$\displaystyle du = dx, v = e^{2x}$

$\displaystyle \int udv = uv-\int vdu$

$\displaystyle xe^{2x}-\int e^{2x}dx$

$\displaystyle xe^{2x}-e^{2x}+C$

what i got is wrong, i think im doing something wrong with $\displaystyle e^{2x}$
• Jun 13th 2007, 10:51 AM
ThePerfectHacker
Quote:

Originally Posted by viet
1) $\displaystyle \int xe^{2x}dx$

Let $\displaystyle u=x \mbox{ and }v' = e^{2x}\Rightarrow u' = 1 \mbox{ and } v= \frac{1}{2}e^{2x}$

Thus,
$\displaystyle \int u v' dx = uv - \int u'v dx$
And hence,
$\displaystyle \int xe^{2x} dx = \frac{1}{2}xe^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}+C$
• Jun 13th 2007, 11:12 AM
Soroban
Hello, viet!

Quote:

I think im doing something wrong with $\displaystyle e^{2x}$ . Yes!
$\displaystyle \int e^{2x}dx \:=\:\frac{1}{2}\,e^{2x} + C$
. . . . . . . .$\displaystyle \uparrow$