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Homework Help: Integral of a fraction

  1. Jun 29, 2008 #1
    2. Relevant equations


    3. The attempt at a solution

    I tried substitution, but that wouldn't work. I tried integration by parts, but I must not have done it properly since my answer was several factors of x off. Am I missing something?
  2. jcsd
  3. Jun 29, 2008 #2
    Use parenthesis!!! I'm assuming your problem is:

    [tex]\int\frac{\sqrt{x^3} +1}{\sqrt x+1}dx[/tex]

    So, where is your work? Thanks :)
  4. Jun 29, 2008 #3
    Please use brackets!

    Now it isn't clear what your integral is. :frown:

    Is it the integral below? Is het root taken only of x^3 and x or the whole term?

    [tex] \int \frac{ \sqrt{x^3}+1}{\sqrt{x}+1} \mbox{d}x [/tex]
  5. Jun 29, 2008 #4
    Assuming rocomath is correct in formulating the problem, try long division before integration
  6. Jun 29, 2008 #5
    Right. Here's my work:

    u=x[tex]^{3/2}[/tex]+ 1 du= [tex]\frac{3}{2}[/tex]x[tex]^{1/2}[/tex]
    dv = x[tex]^{1/2}[/tex] + 1 v= 2x[tex]^{3/2}[/tex] + x

    After that I decided to check the answer before proceeding further to see if I was on the right track, which I wasn't. The books answer was (1/2)x^2 - (3/2)x[tex]\sqrt{x}[/tex] + x + C, and I'm at a loss as to how that happened.
  7. Jun 29, 2008 #6
    I let [tex]x=u^2 \rightarrow dx=2udu[/tex]

    Then I used polynomial division.
    Last edited: Jun 29, 2008
  8. Jun 30, 2008 #7
    divide sqrt(x)+1 into x*sqrt(x)+1, the integration then becomes very straight forward
  9. Jun 30, 2008 #8


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    Quite so, or factor [tex]x^{3/2} + 1[/tex] as a sum of two cubes...
  10. Jun 30, 2008 #9
    Oh! Very nice, didn't even notice that.
  11. Jun 30, 2008 #10
    thanks guys!
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