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Integation by parts

Yuuki

Member
Jun 7, 2013
43
how do i integrate
(xe^(2x))/((1+2x)^2)??
do i substitute 1 + 2x = w?

but if i do, how do i proceed from there?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: integation by parts

I would definitely recommend $w=1+2x$ to begin. Linear substitutions like that cost you nothing, and could gain you quite a bit, as in this case. What is the resulting $w$ integral?
 

Yuuki

Member
Jun 7, 2013
43
Re: integation by parts

i got 1/4(e^(w-1)/w - e^(w-1)/w^2))
but this doesn't look any easier... :/
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: integation by parts

Here, let me type it out $\LaTeX$ style:
$$\frac{1}{4e} \int \frac{e^{w}(w-1)}{w^{2}} \, dw.$$
Is that what you have?
 

Yuuki

Member
Jun 7, 2013
43
Re: integation by parts

yes, and i notice one technique you used, which is to take out the e^(-1), but i still don't see how i should go..

thx for the help btw :)


EDIT:
actually i found the answer;
i had to let u = xe^(2x)
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: integation by parts

Another way to solve this integral is to notice that
\begin{align*}
\frac{d}{dx} \frac{f(x)}{g(x)}&=\frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \\
\frac{f(x)}{g(x)}+C&=\int \frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \, dx.
\end{align*}
The quotient rule doesn't come in handy all that often, but when it does, it surprises you.