# Integral of a Natural Log

1. Feb 5, 2004

### wubie

Hello,

It has been over a year since I last took calculus. And I don't recall how to take the integral of a natural logarithmic function. Here is the question that I am supposed to integrate.

double integral 1/(x+y) dA

where

R = [1,2] X [0,1]

So what I did first was integrate with respect to y first. I ended up with

ln(x+y)

with an upper limit of 1 and a lower limit of 0. Once simplified I get

ln(x+1) - ln x or ln( (x+1)/x )

Now I have to integrate with respect to x. But I can't remember how to take the integral of a natural log function. How do I proceed from here?

I can't remember if I can do the following:

Let G(x) = integral of ln( (x+1)/x ) dx

then

e^G(x) = integral of e^ln( (x+1)/x ) dx

which would simplify to

integral of (x+1)/x dx.

After I get a solution to the above equation I would then take the log of

ln e^G(x) = ln (answer)

to get

G(x) = ln (answer).

Can I do that? I can't remember. If not, how do I proceed from here?

Any help is appreciated. Thankyou.

2. Feb 5, 2004

### NateTG

You can't move the exponential into the integral like that.

That said, IIRC:

$$\int ln(x) dx= x ln(x)-x$$
(You can derive this by using parts)

3. Feb 5, 2004

### wubie

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

I should have known I could have done it by parts. I see that now. Thanks NateTG.

4. Feb 6, 2004

### Hurkyl

Staff Emeritus
And it's a lot easier to integrate $\ln(x+1) - \ln x$ than it is to integrate $\ln (x+1)/x$.

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