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Homework Help: Integral of Natural Log Part II

  1. Feb 5, 2004 #1

    I am still having trouble with taking the integral of the following:

    integral of ln(x+1) dx

    I am trying to do it by parts but I end up getting stuck. I had no problem doing:

    integral of ln x dx

    But I can't seem to get

    integral of ln (x+1) dx

    Let u = ln (x+1) then du = 1/ (x+1)

    Let dv = dx then v = x.

    So then

    integral of ln (x+1) dx =

    ln(x+1)*x - integral of x/(x+1) dx

    I then though that I would integrate the last part by parts again.

    Let u = x then du = dx

    Let dv = 1/(x+1) then v = ln(x+1)

    So then

    integral of x/(x+1) dx =

    x*ln(x+1) - integral of ln(x+1) dx.

    So far I now have

    integral of ln (x+1) dx =

    ln(x+1)*x - [x*ln(x+1) - integral of ln(x+1) dx]

    But then all I end up getting is

    integral of ln (x+1) dx = integral of ln (x+1) dx

    So I am back to where I started. What am I doing incorrectly?

    Any help is appreciated. Thankyou.
  2. jcsd
  3. Feb 5, 2004 #2
    you know the integral of ln[x] and now u want integral of ln[x+1]?
    Is it just me or do I hear x+1 begging to be substituted??
  4. Feb 6, 2004 #3

    What is wrong with me tonight? I can't see anything. >8(

    Thanks stoffer.

  5. Feb 6, 2004 #4
    no prob. we all have our moments
  6. Feb 6, 2004 #5


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    Science Advisor
    Homework Helper

    [tex]\int \ln(x+1) dx[/tex]
    let [tex]u=x+1[/tex]
    then [tex] du=1 dx [/tex]
    so you have
    [tex] \int \ln(u) du [/tex]
    which is:
    now you can apply the formula I gave you in the other thread so you get:
    [tex]u \ln u - u = (x+1) \ln (x+1) - (x+1)[/tex]

    FYI derivation of the formula:

    [tex]\int \ln(x) dx = \int \ln(x) * 1 dx[/tex]
    now by parts we have:
    [tex]du=\frac{1}{x} dx[/tex]
    [tex]\int 1 * \ln(x) dx=\int u dv = uv+C_1-\int v du = [/tex]
    [tex]x \ln x +C_1- \int x* \frac{1}{x} dx=x \ln x +C_1- \int 1dx = x \ln x +C_1- x+C_2=[/tex]
    [tex]x \ln x -x +C[/tex]
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