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Homework Help: I hate Integration by parts

  1. Dec 12, 2004 #1
    I am usually alright once I figure out how to split up the integral into
    u:
    du:
    v:
    dv:
    so i can simply do
    [tex] uv-\int v*du [/tex] but I keep messing up on there I will post some examples if I can find them and if someone could help me that would be great

    [tex] \int (ln(x))^2 [/tex]
     
  2. jcsd
  3. Dec 12, 2004 #2
    Here is what I am trying

    [tex] \int (ln(x))^2 [/tex]

    u: ln x
    du: 1/x
    v: x ln(x)-x
    dv: ln x

    [tex](x ln(x)-x)ln(x)-\int {ln(x)}-1 [/tex]

    [tex](x ln(x)-x)ln(x)-(x ln(x) -x) + x [/tex]
     
    Last edited: Dec 12, 2004
  4. Dec 12, 2004 #3
    I am also trying to figure out u and dv for [tex] \int (x ln x)^2 [/tex] but i figured it would follow in line once i figured out the previous problem
     
    Last edited: Dec 12, 2004
  5. Dec 12, 2004 #4

    Pyrrhus

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    Hey tom,

    For

    [tex] \int x^2 (\ln x)^2 dx [/tex]

    Use [itex] u = (\ln x)^2 [/itex]
     
    Last edited: Dec 12, 2004
  6. Dec 12, 2004 #5

    quasar987

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    You have the right answer, what's the problem?
     
  7. Dec 12, 2004 #6
    I figured it out and kept editing it... thanks though... I am still not sure how I would know to split up the U and dv in the way that I did
     
  8. Dec 12, 2004 #7
    ok I got it down to
    [tex] (x^2 ln(x))(x ln(x) - x) - [ \frac{x^3(3ln(x)-1)}{27}-\frac{x^3}{27} ][/tex]
     
  9. Dec 12, 2004 #8
    out of curisoity is there any rules on how to do [tex] \int \frac{x^2 ln(x)}{27} [/tex] because it was needed for the previous problem and I just used my calculator to come out with the first fraction after the first bracket.
     
  10. Dec 12, 2004 #9

    Pyrrhus

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    That's an integration by parts

    If we rearrange it:

    [tex] \frac{1}{27} \int x^2 \ln x dx [/tex]
     
  11. Dec 12, 2004 #10

    quasar987

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    Unfortunately, I don't think there's any other way than trial an error.
     
  12. Dec 12, 2004 #11

    Pyrrhus

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    Quasar is right, but you will learn to notice which to pick as you make plenty integration by parts.
     
  13. Dec 12, 2004 #12
    thx... what would I do to start [tex] \int tan^{-1} [/tex]
     
    Last edited: Dec 12, 2004
  14. Dec 12, 2004 #13
    oh wait the derivitive of [tex] tan^{-1} = \frac{1}{(x^2+1)} [/tex] So I set tan^-1 as u and 1 as dv... let me see what i get
     
  15. Dec 12, 2004 #14

    Pyrrhus

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    Assuming your mean arctg

    I recommend [itex] u = arctg [/itex]
     
  16. Dec 12, 2004 #15
    I was doing the [tex] \int x sin^{-1} [/tex] and was wondering once you got to
    [tex] x sin^{-1} - \int \frac{x}{\sqrt{1-x^2}} [/tex] how you would integrate [tex] \int \frac{x}{\sqrt{1-x^2}} [/tex] I know the answer is [tex] -\sqrt{1-x^2} [/tex] but how do you get that
     
  17. Dec 12, 2004 #16
    Alright
    I am doing the [tex] \int x^2 e^{5x} [/tex] and I got to here
    assuming u:x^2 and dv: e^(5x)
    [tex] \frac{x^2*e^{5x}}{5} - \frac{2}{5} \int 2x e^{52} [/tex]

    what would I do now?
     
  18. Dec 12, 2004 #17

    Pyrrhus

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    Assuming you mean 5x instead of 52.

    Integrate by part that integral :smile:
     
  19. Dec 12, 2004 #18
    oh ****... this is going to take forever... I am on problem 18 out of 64 and I realized that the problems ahead will require 5 integration by parts
     
  20. Dec 12, 2004 #19
    lol this is what i mean [tex] \int x^5*e^{2x^3} [/tex]
     
  21. Dec 12, 2004 #20
    woah this was werid....

    i was doing [tex] \int x^3*e^{x^2} [/tex] and I got down to

    [tex] 2x^4 e^{x^2}-2 \int x^3*e^{x^2} [/tex] which is the original problem... how would I go about solving this...
     
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