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Can't integrate by parts an integral with a fraction inside

  1. Jun 16, 2012 #1
    1. The problem statement, all variables and given/known data

    In the section of the book of integration by parts, there is an exercise that I don't even know how to tackle anymore. It's this:

    [tex]\int \frac{xe^{2x}}{(1+2x)^2} dx[/tex]


    2. Relevant equations

    [itex]{uv} - {\int v{du}}[/itex]


    3. The attempt at a solution

    [itex]u = x ; [/itex]
    [itex]du = 1;[/itex]
    [itex]dv= e^{2x};[/itex]
    [itex]v = \frac{1}{2}{e^{2x}}[/itex]

    [itex]= \frac{1}{2} {x}{e^{2x}} - \frac{1}{2} {\int e^{2x} {dx}}[/itex]

    But I know is wrong because I no longer have the [itex]{(1+2x)}^2.[/itex]
     
    Last edited by a moderator: Jun 16, 2012
  2. jcsd
  3. Jun 16, 2012 #2

    Mark44

    Staff: Mentor

    Yes, this is wrong. Whatever you choose for u and dv must multiply to the integrand you start with. I haven't worked this problem, but I would start by doing an ordinary substitution first (either u = 2x or u = 2x + 1), and then try integration by parts.

    Using u = 2x, you get
    $$ 1/4 \int{\frac{ue^u~du}{(u + 1)^2}}$$
     
  4. Jun 16, 2012 #3

    SammyS

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    To start with, your formula for integration by parts is incomplete. It is:
    [itex]\displaystyle \int u\,dv={uv} - {\int v{du}}[/itex] ​
    Then, as you noted, you left part of the integrand out. In other words, you haven't split-up the original integral into u and dv .

    By the way, if [itex]u = x\,, [/itex] then [itex]du = dx\,[/itex]

    and if [itex]dv= e^{2x}dx[/itex] then [itex]\displaystyle v=\frac{1}{2}{e^{2x}}\,.[/itex]

    (I see Mark44 beat me to the punch.)
     
  5. Jun 16, 2012 #4
    I tried that too, I just didn't wrote it in the attempts. Although I forgot about the 1/4 in front of the integral.


    The problem is that this is the same but with other letters. I still don't know how can I possibly integrate and take the denominator into account. I thought about expanding in a sum of integrals and I can integrate the first 1 of them, but I have a problem with the other 2.


    [itex](1+u)^2 =[/itex] [itex]1[/itex] [itex]+[/itex] [itex]2u[/itex] [itex]+[/itex] [itex]u^2[/itex]

    [itex]\frac{1}{4}\int \frac{ue^{u}}{1}[/itex] [itex]+[/itex] [itex]\frac{1}{4}\int \frac{ue^{u}}{2u}[/itex] [itex]+[/itex] [itex]\frac{1}{4}\int \frac{ue^{u}}{u^2}[/itex]


    Then, how can I deal with the 2nd (Blue) and the 3rd one (Green)?

    Ah thanks, I totally forgot the dx.
     
    Last edited by a moderator: Jun 16, 2012
  6. Jun 16, 2012 #5

    vela

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    You can't do that.
    $$\frac{a}{b+c+d} \ne \frac{a}{b} + \frac{a}{c} + \frac{a}{d}$$ You also left the du's out of all the integrals.
     
  7. Jun 16, 2012 #6
    After you get ¼∫xex/(x+1)2 dx (replacing the u's with x's), try u = xex and dv = 1/(x+1)2 dx
     
  8. Jun 16, 2012 #7
    I think I got it with that. Can you correct me?

    [itex]v = ue^u[/itex] ; [itex]dv = e^u(u+1)du[/itex]

    [itex]dw = \frac{1}{(u+1)^2}du[/itex] ; [itex]w = -\frac{1}{(u+1)}[/itex]​

    [itex]= -\frac{ue^u}{(u+1)} + \frac{1}{4} \int \frac{e^u(u+1)}{(u+1)} du[/itex]

    [itex]= -\frac{ue^u}{(u+1)} + \frac{1}{4} \int e^u du[/itex]

    [itex]= -\frac{ue^u}{(u+1)} + \frac{1}{4} e^u + C[/itex]

    So, placing back everything with [itex](u = 2x)[/itex]

    [itex]\int \frac{xe^{2x}}{(1+2x)^2} dx = -\frac{1}{4}\frac{(2x)e^{2x}}{(2x+1)} + \frac{1}{4} e^{2x} + C[/itex]
     
    Last edited: Jun 16, 2012
  9. Jun 16, 2012 #8
    That's right, except for one u that didn't get replaced by 2x :wink:
     
  10. Jun 16, 2012 #9
    You almost got it.

    That 1/4 term though needs to be multiplied to your first term. Remember how Mark44 wrote it - 1/4 * Integral Stuff. What you did, the substitution and integration by parts, is just the evaluation of the "Integral Stuff" in the equation 1/4 * Integral Stuff. So that 1/4 still needs to be appended.

    You can verify your answer using wolfram alpha http://www.wolframalpha.com/input/?i=integrate+x*e^(2x)+/+(1+2x)^2
     
    Last edited by a moderator: Jun 16, 2012
  11. Jun 16, 2012 #10

    Mark44

    Staff: Mentor

    Psinter, Please don't use those HTML SIZE tags. What you write shows up perfectly well without them.
    If you want you LaTeX stuff to be a bit larger use [ tex ] tags instead of [ itex ] (without the spaces).

    You can also use [noparse]$$ <expression> $$[/noparse], which is the same as [ tex ] and [ /tex ].

    Or you can use [noparse]## <expression> ##[/noparse], which is the same as [ itex ] and [ /itex ].
     
  12. Jun 16, 2012 #11
    Thanks, fixed that.
    Fixed too. :smile:

    Got it. It's just that in my eyes I almost couldn't see the e exponent. Thanks for those, I'll use them next time.

    This isn't an exercise for a work, but it took me so long to do it I got fond of it and made a comic: (Wolfram got no steps, but I wanted the steps to learn how to do it)
    kBP6Q.jpg
     
  13. Jun 16, 2012 #12

    Mark44

    Staff: Mentor

    Cool!
    What does le' mean?
     
  14. Jun 16, 2012 #13
    It's French. The translation would be "the". However, in Enlgish is used as an expression when mocking something. It's pronounced like "e" sounds in Japanese, not like "e" sounds in English.

    That kind of meme comic has always used "le" to refer to the one who is being talked about so I thought I should use it too, just for tradition. I added the apostrophe to sort of separate it from being pronounced together with the next word but in tradition it doesn't have that apostrophe.
     
  15. Jun 16, 2012 #14
    Just one last note: sometimes you need to change the problem so Wolframalpha likes it and shows you its steps. For some reason WA didn't like the e2x in the problem, but entering the new integral after doing the substitution u = 2x instead will show you the steps for that integral!
     
  16. Jun 17, 2012 #15
    Nice. Thanks for that one.
     
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