Integral of Natural Log Part II

[wubie]

Hi,

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.

 

[stoffer]

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??

 

[wubie]

!@#$!@#$!@#$!@#$%!@#$!

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

Thanks stoffer.

Cheers.

 

[stoffer]

no prob. we all have our moments

 

[NateTG]

[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]u=\ln(x)[/tex]
[tex]dv=1dx[/tex]
so
[tex]du=\frac{1}{x} dx[/tex]
[tex]v=x[/tex]
so
[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]