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Integral ∫x^3*√(x^2-5) dx

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data

    ∫x^3*√(x^2-5) dx

    2. Relevant equations

    ∫u.dv=u.v-∫du.v


    3. The attempt at a solution

    So i tried to change the integral to ∫x*x^2*√(x^2-5)dx and u = x^2-5, then du = 2x, so 1/2*∫x^2*√(x^2-5) . Let u = √(x^2-5) , du = x/√(x^2-5) and dv = x^2 , v = x^3/3. Am I going in the right direction?

    Thanks in advance
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Nov 25, 2012 #2

    Dick

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    Re: Integral

    You don't need parts. Your first substitution is the one to use, u = x^2-5. Notice that means x^2=5+u.
     
  4. Nov 25, 2012 #3
    Re: Integral

    I did that, u=x^2-5, du= 2xdx, x^2=u+5 and I got :

    1/2*∫(u+5)*√udu

    so I multiplied and I got : 1/2*∫u^(3/2)+5u^(1/2)

    1/2*1/5 * ∫u^(3/2)+u^(1/2)

    and I got u^(5/2)/25+(2u^(3/2))/3 and It's wrong :/ where did I failed?

    Thanks
     
  5. Nov 25, 2012 #4

    Dick

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    Re: Integral

    Just sloppy algebra, I think. Where did the '1/5' come from?
     
  6. Nov 25, 2012 #5
    Re: Integral

    To remove the 5 from 5u^(1/2)

    The solution is supposed to be:

    x^3/3*(u^(3/2))-2/15*(u^(5/2))+c
     
  7. Nov 25, 2012 #6

    Dick

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    Re: Integral

    'Remove the 5'?? I don't get it. And the given solution looks wrong as well.
     
  8. Nov 25, 2012 #7
  9. Nov 25, 2012 #8

    Dick

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  10. Nov 25, 2012 #9
    Re: Integral

    So where is my mistake?
     
  11. Nov 25, 2012 #10

    Dick

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    Re: Integral

    Start with 'removing the 5'. You can write u^(3/2)+5u^(1/2)=5*((1/5)*u^(3/2)+u^(1/2)). That's not (1/5)*(u^(3/2)+u^(1/2)), if that's what you meant. It's hard to tell how you are grouping things. Use more parentheses to make things clearer.
     
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