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Integral help

  1. Apr 3, 2010 #1
    Is this integral doable at all ?

    x^4 dx / (x^2+a^2)^(3/2)

    I have tried several u-substitutions but none of them seem to make the form of the integral any simpler
  2. jcsd
  3. Apr 3, 2010 #2


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    I looks doable to me. Did you try a trig substitution like x=a*tan(t)?
  4. Apr 4, 2010 #3
    Thanks :)
    so this is how far I get:

    if x=atan(u)
    then dx=asec^2(u)du
    my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

    since (x^2+a^2)^1/2 = asec(u)
    then my denominator will be a^3sec^3(u)

    my old limits of integration were x=0 to x=R
    new ones become u=0 to u=tan^-1(R/a)

    so my new integral is:

    integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

    unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either
  5. Apr 4, 2010 #4


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    Now try v=sin(u) and remember tan(u)=sin(u)/cos(u). I said it was doable. I didn't say it was easy.
  6. Apr 4, 2010 #5
    So , if I let v=sinu, then dv=cosudu ---> du=dv/cosu >> tan^4ucosu = v^4dv /cos^4(u) = v^4 dv /(1-v^2)^2

    Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?
  7. Apr 4, 2010 #6


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  8. Apr 4, 2010 #7
    Wow thank you so much!

    That was one hell of an integral..!
  9. Apr 4, 2010 #8
    x = a sinh(t) is a lot simpler. With some effort you can see the answer using mental computations, without using paper and pencil.
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