Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration problem

  1. Feb 24, 2005 #1
    I cannot figure out how to integrate (X/X+d)dX. I know that I should probably use the method of partial fractions but the num and den degrees are the same and the den cannot be broken down. I guess I could use some sort of substitution but which? please help :confused:
     
  2. jcsd
  3. Feb 24, 2005 #2
    try using long division to divide x by (x+d) & see what you get.
     
  4. Feb 24, 2005 #3
    I'd do integration by parts.

    You can write this as [tex]x * \frac{1}{x+d} dx[/tex]

    The second part should look familiar. (arctan)

    The [tex]\int{u*dv} = uv - \int{v*du}[/tex]

    Let u = x and dv = [tex]\frac{1}{x+d}[/tex]

    Edit: I made a mistake. Sorry. For the arctan rule to apply there needs to be an x^2 in the bottom. Maybe you can still do it that way. I don't know.
     
    Last edited: Feb 24, 2005
  5. Feb 24, 2005 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What does that have to do with anything?
     
  6. Feb 24, 2005 #5
    wait... if I let u = x+d then x = u-d and int (x/x+d)dx becomes
    int(1-d/u)du = u - dlnu = x+d - dln(x+d). Is this correct??

    You're very helpful :approve:
     
  7. Feb 25, 2005 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks, I try. :smile:

    I was trying to hint that partial fractions still applies when the numerator and denominator have the same degree. (At least, the first step of long division does)
     
  8. Feb 25, 2005 #7
    Very Simple Way, Here It,

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

    then the answer will be:

    x - d*ln(x+d) + C

    hehe, isn't that easy?????
     
    Last edited: Feb 25, 2005
  9. Feb 25, 2005 #8
    The above is correct, except it is "-" instead of "+".
     
  10. Feb 25, 2005 #9
    HEHE, right, i fixed it :P
     
  11. Feb 25, 2005 #10
    Ok, I see, thanks. However, I don't understand why the below procedure doesn't work when it's just another simple substitution??

    if I let u = x+d then x = u-d and int (x/x+d)dx becomes int(1-d/u)du = u - dlnu = x+d - dln(x+d).
     
  12. Feb 25, 2005 #11

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because you forgot the constant of integration.
     
  13. Feb 25, 2005 #12
    I understand that I forgot the constant of integration, but I am referring to the fact that Destroyer's procedure produces: x - dln(x+d) + C while the other procedure produces (x+d) - dln(x+d) + C. Shouldn't they produce the same value considering they are merely two different substitutions?
     
  14. Feb 25, 2005 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The two answers are the same.
     
  15. Feb 25, 2005 #14
    no ... x does not equal x+d

    you're not helping me here
     
  16. Feb 26, 2005 #15

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes he is...he told you the answers were the same. You should have believed him. x is not x + d. But what is d? d is a constant. So did you really get a different answer from TheDestroyer's (lol at the name)? No, you did not. Both answers are valid *antiderivatives*, because when you differentiate them with respect to x, all constant terms drop out...and you end up back with the integrand. So absorb d into C...they are both constant terms. Your new C = C+d. Happy? :wink:
     
  17. Feb 26, 2005 #16
    Ok wait, I just realized I made a mistake at the beginning of the thread. I should have clarified that this isn't an indefinite integral. The limits are from (0 to L) so:

    1) let u = x+d then x = u-d
    (x/x+d)dx = (1-d/u)du = u - dlnu = x+d - dln(x+d)

    from (0 to L) = (L+d) - dln(1+L/d)

    2) (x)/(x+d)dx = (x+d-d)/(x+d)dx = (1) - (d/(x+d))dx = x - dln(x+d)

    from (0 to L) = L - dln(1+L/d)

    am I still missing it, or are they still the same?

    PS. forgive for going on and on :blushing:
     
    Last edited: Feb 26, 2005
  18. Feb 26, 2005 #17
    Yes, they are still one and the same. When taking the definite integral, you will subtract one from the other, so:

    [tex]g(L) - g(0) = L + d - d * \ln(L+d) - (0+d - d * \ln(0+d))
    = L - 0 + d - d - d * \ln(L+d) + d * \ln(0+d) = L - d(\ln(L+d) - \ln(d))
    = L - d * \ln(L/d + 1)[/tex]

    Notice how the d cancels out the -d (just like the arbitrary constant of integration, C, cancels out) when you take the definite integral.
     
    Last edited: Feb 26, 2005
  19. Feb 26, 2005 #18
    holy crap ... lol. I can't believe I didn't see that :surprised
    thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration problem
  1. Integral problem (Replies: 3)

  2. Problem with integration (Replies: 16)

  3. Integration Problem (Replies: 1)

  4. Integral problem (Replies: 2)

  5. Integral Problem (Replies: 1)

Loading...