1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook