integration by parts

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

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

$\int udv = uv-\int vdu$

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

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

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

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

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

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

Quote:

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