How to integrate [ x / (1 + x) ] dx

  1. Jul 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Evalute the integral

    ∫ [x / 1 + x] dx


    2. Relevant equations
    ∫ [x / 1 + x] dx


    3. The attempt at a solution
    I forgot how to do solve this type of integral, or never had enough practice. And this problem is actually for a physics problem :-) And my algebra is very rusty as well.

    ∫ [x / 1 + x] dx
    Can't do U-sub, don't think you can do by parts, can't do trig substitution, not sure about partial fractions

    Divide everything by 'x'.
    ∫ [x / 1 + x] dx = ∫ [x (1 / x)/ 1(1 / x) + x(1 / x)] dx

    ∫ [1 / (1/x) + (1)] dx

    Not sure where to go from here.
     
  2. jcsd
  3. Jul 23, 2011 #2

    LCKurtz

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    Try writing

    [tex]\frac x {1+x}=1 -\frac 1 {1+x}[/tex]
     
  4. Jul 23, 2011 #3
    Yes, I see that will work, but I just can not see a straightforward way to get there :-(
     
  5. Jul 23, 2011 #4

    SammyS

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    ... to get there from where?

    Can you find [itex]\displaystyle \int\,\left(1 -\frac 1 {1+x}\right)\,dx\ ?[/itex]
     
  6. Jul 23, 2011 #5

    LCKurtz

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    It is a standard procedure when you have a polynomial of equal or higher degree in the numerator to reduce it by long division until the numerator is less degree than the denominator. Just divide x+1 into x long division.
     
  7. Jul 23, 2011 #6
    There's a trick that's often used for problems like this:

    [tex]\frac{x}{x + 1} = \frac{x + 1 - 1}{x + 1} = \frac{x + 1}{x + 1} - \frac{1}{x + 1} = 1 - \frac{1}{x + 1}[/tex]

    You could also do this easily with the substitution u = x + 1 → u - 1 = x and du = dx. After the substitutions, then you can split up the integrand into two fractions, which could be a little easier than the above method.
     
  8. Jul 23, 2011 #7

    SammyS

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    Let's wait to hear from Cyan before we do any more. We're not sure what his/her hang up is.
     
  9. Jul 24, 2011 #8
    My main point:
    The word easy should not be in a tutor's vocabulary. Although, for those who have spent hours (sometimes hundreds) mastering various Calculus material, few if any of us thought it was easy to learn.


    The most effective method, for most students is to realize, not only u=x+1 can be substituted in, but we may replace other x's using x=u-1. I hope this didn't get lost in the mix.


    I hope SammyS is Ok, with this interjection.:uhh:
     
  10. Jul 26, 2011 #9
    :cry:

    Thanks, this really helps a lot!

    The only problem is that I get a slightly different answer for the U-Substitution method you provided, probably because I'm being dumb making mistakes here and there:

    ∫ [ x / (x + 1) ] dx

    u = x + 1 ; du = dx

    x = u - 1

    ∫ [ (u - 1) / u ] du

    = ∫ [ u / u] du - ∫ [1 / u] du

    = ∫ 1 du - ∫ [1 / u] du

    = (u) - ln(u) ; recall u = x + 1; = x+1 - ln (x+1)


    The first method you said (and I'll copy it over):

    x / (x + 1) = (x + 1 - 1) / ( x + 1);

    = [ (x + 1) / (x + 1) ] - [ 1 / (x + 1) ]


    = 1 - [1 / (x + 1)]

    Integrating this now:

    ∫ 1 - [1 / (x + 1)] dx

    = x - ln(x + 1)
     
  11. Jul 26, 2011 #10
    Two things for *indefinite integrals*, always remember the constant of integration. Get all the easy points on quizzes & tests.

    Second, they only differ by a constant, right?
    So pick different constants of integration to compensate. +4 for one & +5 for the other.
    Very commonly when doing an integral by two different methods, the answer will come out the same, but differ by an addition constant.

    Thank you very much for coming back and finishing the question. Next step for you: develop more self sufficiency on questions like this.
     
  12. Jul 26, 2011 #11
    Yeah I should have put the + C also, but this was for a physics problem and this integral was the last step from me finishing it, and it actually was a definite integral where I integrated over a length.
     
  13. Feb 19, 2013 #12
    I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?
     
  14. Feb 19, 2013 #13
    I realize that this is a ridiculous question, but what algebraic technique are you using? Are you completing the square?
     
  15. Feb 19, 2013 #14

    SammyS

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    I doubt the technique has a formal name.

    You can get same result by using long division.
     
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