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Homework Help: Integration- is it possible to integrate this?

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the integral of x over ((1+x)1/2-(1+x)1/3) dx


    2. Relevant equations

    integration by substitution.

    3. The attempt at a solution

    I have tried the substitution u = (1+x)5/6 but not really sure if this makes it any better!
    du/dx = 5/6 * (1+x)-1/6

    6/5 * du = (1+x)-1/6 dx

    (1+x) = u6/5

    x= u6/5 - 1

    Take a factor of (1+x)1/6 of the denominator and rewrite to give:

    6/5 * integral of (u6/5 - 1)/(u2/5 - u1/5) du

    Is is possible to integrate this? Or do I need a completely different substitution?
    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2010 #2

    D H

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    Staff Emeritus
    Science Advisor

    What made you pick that??

    Try the much simpler u=(1+x)1/6 (alternatively, u6=1+x).
     
  4. Sep 20, 2010 #3
    Hmmn I was trying to get something so that the integral du/dx was a factor. I thought I'd already tried and failed with (1+x)^1/6 but will try again! Thanks.
     
  5. Sep 20, 2010 #4
    Ok soo here goes:
    u = (1+x)1/6

    du/dx = 1/6 * (1+x)-5/6

    6du = (1+x)-5/6 dx

    u6 = (1+x)

    x = u6 -1

    So integral of x over (1+x)-5/6 * ((1+x)-1/3 - (1+x)-1/2)dx

    Subst for u to give:
    6 * integral of (u6 - 1) / (u-2 - u-3) du

    = 6 * integral of (u9 - u3) / (u - 1) du


    From here I made a second substitution of z = u - 1 (du = dz) which gives
    6 * integral of ((z+1)9 - (z+1)3) / z dz


    Using binomial theorem I get:

    (2/3 * z9) + (27/4 * z8) + (216/7 * z7) + 84z6 + (756/5 * z5) + 189z4 + 166z3 + 99z2 + 36z


    where z = ((1+x)1/6 -1)

    Not exactly the neatest answer =/
     
  6. Sep 20, 2010 #5

    D H

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    Science Advisor

    Rather than making another substitution, why don't you just simplify (u9-u3)/(u-1)?

    Hint: What is (u6-1)/(u-1)?
     
  7. Sep 20, 2010 #6
    Yep I'm being very slow! Thanks should be able to finish now. Thought that answer looked ridiculously long.
     
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