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Homework Help: Antiderivative of the following function

  1. Sep 25, 2005 #1
    How would you guys come about getting the antiderivative for the following function?

    8u^5 / ((4 + u^4)^2)

    I tried rearranging the equation like the following:

    8u^5 * ((4 + u^4)^(-2)), but it's not really getting me anywhere. I tried to do a u substitution, but can't find one where it knocks off the top u^5.

    Thanks in advance.
     
  2. jcsd
  3. Sep 25, 2005 #2
    I take it that you're taking the A.D. wrt u?
    My TI89 gives:
    pi*atan(u^2/2)/180 - (2*u^2/(u^4+4))
    Which in turn tells me that your calculus text should have a table for it in one of the appendices.
    Look for it in the TOC, or index.
    Look for an indefinite integral that has the general form:
    a*x^(n)/(x^(n-1) +(n-1))^(2)
    Or something similar.
     
  4. Sep 25, 2005 #3
    Integration by parts and you'll have to use trig substitution. That's a nasty one. Good luck!
     
  5. Sep 26, 2005 #4

    hotvette

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    Here are a couple of hints. Try the substitution x = u2. That should help reduce things to a more manageable integrand. After that, a little clever algebra and a trig substitution will get you there. Not a trivial problem.

    hotvette
     
    Last edited: Sep 26, 2005
  6. Sep 29, 2005 #5

    Hmm, any hint on what the clever algebra should be? I can't really figure it out from here. I have simplified the equation down to

    4x2 / (4 + x2)2 dx

    using x = u2
     
  7. Sep 29, 2005 #6

    hotvette

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    Try adding and subtracting the same number from the top. Gather terms and split the problem into 2 problems.

    Example: if I have y in the numerator, y - 2 + 2, or (y - 2) + 2 is the same thing. Just pick the right number to add and subtract.
     
  8. Sep 29, 2005 #7
    Thanks for the reply. Am I on the right track?


    4x2 / (4 + x2)2 dx
    = (4x2 + 16 - 16) / (4 + x2)2 dx
    = 4((x2 + 4) - 4)) / (4 + x2)2 dx
    = (4 / (4 + x2)) - (4/ (4 + x2)2) dx
     
  9. Sep 29, 2005 #8

    hotvette

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    You bet. The first one is readily integrated and the second is w/ a trig substitution. My college calc book has an entire section on appropriate trig substitutions based on what the integrand looks like - that's where I found the appropriate trig substituion.
     
  10. Sep 29, 2005 #9
    Thanks. However, can you spot my mistake when solving for the antiderivative of (4/ (4 + x2)2)? I must be doing something wrong, but I can't find what

    (4/ (4 + x2)2)

    Let x = 2 tan theta
    dx = 2 sec2theta dtheta

    = (8 sec2theta / ((4 + 4tan2theta)2)

    factor out a 2

    = 4 + 4 tan2 theta / ((4 + 4tan2theta)2)
    = 1 / (4 + 4tan2theta)
    = tan inverse (tan theta)
     
  11. Sep 29, 2005 #10

    hotvette

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    The part I don't follow is how you get from:

    [tex]\frac {d \Theta}{(4 + 4 tan^2 \Theta)}[/tex]

    to

    [tex]tan^{-1} tan \Theta[/tex]

    By the way, I had a slightly different integrand, with 16 in the numerator instead of 4, but frankly, I don't want to take the time to re-check it. Mine might be wrong, I don't know. My objective here is getting you going on the right path, not necessarily verifying the exact answer.
     
  12. Sep 29, 2005 #11
    Alright, thanks. I'll just try it again.
     
  13. Sep 29, 2005 #12
    I think there's a problem in this substitution.

    [tex]\int \frac{8u^5}{(4+u^4)^2}du[/tex]

    [tex]u=x^2[/tex]
    [tex]du=2xdx[/tex]
     
  14. Sep 29, 2005 #13

    hotvette

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    No. The substitution is [itex]x=u^2[/itex]. You have it backwards.

    [tex]\int \frac{4x^2}{(4+x^2)^2} \ dx[/tex]

    is correct after the substitution.
     
  15. Sep 29, 2005 #14

    hotvette

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    I just checked my work again. Here's what I get:

    [tex]\int \frac{4x^2}{(4+x^2)^2} \ dx \ = \ \int \frac{4}{(4 + x^2)} \ dx \ - \int \frac{16}{(4 + x^2)^2} \ dx [/tex]

    Btw, it might be prudent to reduce even further by factoring out the 4's to make it (1 + something^2) instead of (4 + x^2). May not be critical, though.
     
    Last edited: Sep 30, 2005
  16. Oct 4, 2005 #15
    Yup, that's what I got after the substitution and working. Sorry for replying so late. I finally had a chance to do this question just now.


    [tex] \int \frac{16}{(4 + x^2)^2} \ dx = \frac{32sec^{2}\Theta}{(4 + 4tan^{2}\Theta)^2} [/tex] d theta

    where [tex] x = 2tan\Theta[/tex]
    [tex] \ dx = 2 sec^{2}\Theta[/tex] d theta

    After all that, the result I get is [tex]\Theta + \frac{sin2\Theta}{2}[/tex], but for some reason, when I use my calculator to see if this answer is correct, it doesnt match up with the derivative values of the original equation 8u^5 / (4 + u^4)^2
     
  17. Oct 4, 2005 #16

    hotvette

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    Don't forget the first integral.
     
  18. Oct 4, 2005 #17
    Yup, I checked it with both the first and the second integral and no dice. The antiderivative for

    [tex]\int \frac{4}{(4 + x^2)} \ dx[/tex]

    is [tex]2tan^{-1}(x/2)[/tex]
     
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