integration by parts

**viet** · June 13th 2007, 10:38 AM

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 $e^{2x}$

**ThePerfectHacker** · June 13th 2007, 10:51 AM

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,
$\int xe^{2x} dx = \frac{1}{2}xe^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}+C$

**Soroban** · June 13th 2007, 11:12 AM

Hello, viet!

I think im doing something wrong with $e^{2x}$ . Yes!

$\int e^{2x}dx \:=\:\frac{1}{2}\,e^{2x} + C$
. . . . . . . . $\uparrow$

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