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Integral questions

  1. Feb 7, 2004 #1
    Can anybody help me with the integral of 1/ (x^1/3 + x^1/4)
    (cube root and fourth root of x) I dont really know where to start.

    Also my roomate and i were wondering if x^sin(x) exists or if it has to be expressed and integrated as some sort of series.(something i havent learned yet)
     
  2. jcsd
  3. Feb 7, 2004 #2
    Perhaps you could rewrite it as 1 / (x^(1/4) * (1 + x^(1/12))) and use partial fractions? I don't really know. Looks like it'll be pretty messy...
     
  4. Feb 7, 2004 #3

    matt grime

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    try substituting x = y^12

    as for x^sin(x).. it's exp{log(x^sinx)} = exp{sinx*log(x)}

    that's the standard way of defining x^f(x)
     
  5. Feb 7, 2004 #4
    ok thanks for your help guys
     
  6. Mar 14, 2007 #5
    just take the ln of that.. and do the rest intuitively, but hey what do i know, im only 16.
     
  7. Mar 22, 2007 #6
    That is certainly not the same as the function in the original post.
     
  8. Mar 22, 2007 #7
    That still isn't the same function. You can't manipulate fractions like that, it just doesn't work.
     
  9. Mar 22, 2007 #8

    morphism

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    This is why I'm not a fan of rushing people into calculus without a solid foundation in the basics of algebra. (In reference to d_leet's quotes.)

    Anyway, matt's substitution kills this integral. You can also re-write the integrand as:

    [itex]\frac{x^{12}}{x^4 + x^3} = \frac{x^9}{1 + x} = \frac{x^9 + 1 - 1}{1+x}[/itex]

    Then proceed...
     
  10. Mar 23, 2007 #9

    dextercioby

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    Actually it's a power less in the numerator:

    [itex] I=\int \frac{dx}{\sqrt[3]{x}+\sqrt[4]{x}}=12 \int \frac{y^{8}}{y+1}{}dy [/itex]

    ,where [itex] y=\sqrt[12]{x} [/itex].
     
  11. Mar 23, 2007 #10

    morphism

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    I didn't use the substitution...
     
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