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Weird integral

  1. Dec 9, 2003 #1
    How do I do this
    \int x\sqrt[3]{x-1}\,dx
  2. jcsd
  3. Dec 9, 2003 #2
    Seems like a good candidate for integration by parts.

    Let u=x and dv=(x-1)^1/3
  4. Dec 9, 2003 #3
    I assume that's x3 * √(x-1) [ and not x * (x-1)^(1/3) ].

    You can use integration by parts (3 times).

    (There may be an easier way, but I don't see it.)
  5. Dec 10, 2003 #4
    U can substitute [tex] x-1=t^2[/tex]
  6. Dec 10, 2003 #5
    U can substitute [tex] x-1=t^2[/tex]
  7. Dec 10, 2003 #6
    Actually it is x(x-1)^(1/3), So I hope all this works for my problem. Actually, with whatever I've learned so far my teach says it can't be done, I was just curious if it actually can.
  8. Dec 10, 2003 #7


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    Let u= x-1. Then x= u+1 and dx= du so the integrand is

    (u+1)u1/3du= (u4/3+ u1/3)du and it's easy.
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