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September 1st, 2014, 07:39 AM   #1
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Integration by Parts Help

Hello everyone,

This is my first time in this forum, but like other forums online I am sure there is an amazing community of people out there willing to help others; at least that's my personal experience; I know it will be similar in this forum.

I am starting Integration by parts in my Calculus 2 class and it's something that I am still working on to grasp (it's tricky).

I have this exercise and I cannot move forward (it might seem basic I know).

$\displaystyle \int \frac{xe^{2x}}{(1+2x)^{2}}dx$

I chose:
$\displaystyle u=e^{2x}$
$\displaystyle du=2e^{2x}dx$
$\displaystyle v=\frac{x}{(1+2x)^{2}}$
$\displaystyle dv=?$

But I cannot get the integral of dv to find v.

Any ideas?

Thanks a lot.

Last edited by skipjack; September 2nd, 2014 at 02:49 PM.
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September 1st, 2014, 07:44 AM   #2
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Before trying the integration by parts I would try substituting $\displaystyle u = 1 + 2x$. It simplifies the integral
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September 1st, 2014, 08:05 AM   #3
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Thanks for the replay Benit,

I tried substituting $\displaystyle u=1+2x$ first in the original integral but the result is not that easy to work with ( or I am not seeing it)



$\displaystyle \frac{1}{4}\int\frac{e^{u-1}\left ( u-1 \right )}{u^{2}}du$

Last edited by fredlo2010; September 1st, 2014 at 08:08 AM.
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September 1st, 2014, 12:40 PM   #4
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Quote:
Originally Posted by fredlo2010 View Post
Thanks for the replay Benit,

I tried substituting $\displaystyle u=1+2x$ first in the original integral but the result is not that easy to work with ( or I am not seeing it)



$\displaystyle \frac{1}{4}\int\frac{e^{u-1}\left ( u-1 \right )}{u^{2}}du$
$\displaystyle \int\frac{u-1}{u^2}du = \ln(u)+\frac{1}{u}$

It may not help.

Last edited by skipjack; September 2nd, 2014 at 02:59 PM.
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September 1st, 2014, 03:45 PM   #5
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$\displaystyle u=xe^{2x},\,du=e^{2x}(1+2x)\,dx$

$\displaystyle dv=(1+2x)^{-2}\,dx,\,v=-\frac12(1+2x)^{-1}$
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September 1st, 2014, 05:15 PM   #6
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle u=xe^{2x},\,du=e^{2x}(1+2x)\,dx$

$\displaystyle dv=(1+2x)^{-2}\,dx,\,v=-\frac12(1+2x)^{-1}$
Hi Greg,

Thanks a lot for the reply. That did the trick.

This is the final answer. I am not sure if it's correct to post all the solution? Can you guide me. In other forums we do to help other users in case they face a similar problem (programming stuff) Also I do not see a "Solved" flag. My guess is that it's not used.

$\displaystyle \frac{e^{2x}}{8x+4}+C$

Thanks to all of you for the help.

Last edited by skipjack; September 2nd, 2014 at 02:52 PM.
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September 1st, 2014, 05:59 PM   #7
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There's no harm in posting the solution, though we do encourage others to show some effort towards solving the problem (as you have).
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September 1st, 2014, 06:18 PM   #8
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Thanks Greg,

Here is the worked out solution. I hope its legible enough

Solution.jpg

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September 2nd, 2014, 12:49 PM   #9
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Quote:
Originally Posted by mathman View Post
$\displaystyle \int\frac{u-1}{u^2}du =
\ln(u)+\frac{1}{u} \ \ \ \ $ <-------

It may not help.
That should be:

$\displaystyle \int\frac{u-1}{u^2}du \ = \ \ln|u| \ + \ \frac{1}{u} \ + \ C$

Last edited by skipjack; September 2nd, 2014 at 02:57 PM. Reason: ln -> \ln
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