My Math Forum Integration by Parts Help
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 September 1st, 2014, 07:39 AM #1 Newbie   Joined: Sep 2014 From: Miami Posts: 18 Thanks: 2 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.
 September 1st, 2014, 07:44 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,142 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Before trying the integration by parts I would try substituting $\displaystyle u = 1 + 2x$. It simplifies the integral
 September 1st, 2014, 08:05 AM #3 Newbie   Joined: Sep 2014 From: Miami Posts: 18 Thanks: 2 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.
September 1st, 2014, 12:40 PM   #4
Global Moderator

Joined: May 2007

Posts: 6,705
Thanks: 670

Quote:
 Originally Posted by fredlo2010 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.

 September 1st, 2014, 03:45 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond $\displaystyle u=xe^{2x},\,du=e^{2x}(1+2x)\,dx$ $\displaystyle dv=(1+2x)^{-2}\,dx,\,v=-\frac12(1+2x)^{-1}$ Thanks from fredlo2010
September 1st, 2014, 05:15 PM   #6
Newbie

Joined: Sep 2014
From: Miami

Posts: 18
Thanks: 2

Quote:
 Originally Posted by greg1313 $\displaystyle u=xe^{2x},\,du=e^{2x}(1+2x)\,dx$ $\displaystyle dv=(1+2x)^{-2}\,dx,\,v=-\frac12(1+2x)^{-1}$
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.

 September 1st, 2014, 05:59 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond There's no harm in posting the solution, though we do encourage others to show some effort towards solving the problem (as you have).
 September 1st, 2014, 06:18 PM #8 Newbie   Joined: Sep 2014 From: Miami Posts: 18 Thanks: 2 Thanks Greg, Here is the worked out solution. I hope its legible enough Solution.jpg Thanks Thanks from greg1313
September 2nd, 2014, 12:49 PM   #9
Banned Camp

Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

Quote:
 Originally Posted by mathman $\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

 Tags integration, parts

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post devinperry Calculus 1 April 14th, 2014 01:26 PM Zynoakib Calculus 2 March 14th, 2014 05:21 PM cookster Calculus 4 June 2nd, 2012 05:10 AM jakeward123 Calculus 16 February 24th, 2011 07:34 AM tnutty Calculus 1 February 23rd, 2009 01:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.