Integrate ln(1+x) dx between 2 and 0

[Monkeyseat]

Hi,

Question

Integrate ln(1+x) dx between 2 and 0.

Working

Okay, so I know you have to use integration by parts (leting u = ln(1+x) as ln(1+x) cannot be integrated directly):




So now you have to use integration by parts again:




This brought me back around in a full circle to what I was originally trying to integrate. If I used integration by parts again I'd keep going around in a circle. I tried swapping the 'variables' over when integrating by parts for a second time but got the same problem:




I'm stuck - I don't how to finish this to integrate ln(1+x) dx. How do I integrate (x/x+1) using integration by parts??? Is what I have done so far ok? Maybe a different approach is required, I don't know whether it is possible to integrate it by parts without going around in circles.

By the way, I know (x/x+1) can be split up to ((x+1)-1)/(x+1) and then 1 - (1/(x+1)) and integrated to give [x - ln(1+x)], but I just want to know if it's possible to do it by parts, without getting stuck in a 'loop'. I don't think I've made a mistake and I can't see where to go with this. I think I am only at a 'basic' level using integration by parts at the moment (not sure how much farther it goes), so I may not have studied any 'advanced' techniques needed to do this yet.

Just wondering.

Thanks for any help.

 

[galactus]

A simple u substitution will do this lickety split. Do you HAVE to use parts?.

\(\displaystyle \int_{0}^{2}ln(x+1)dx\)

Let \(\displaystyle u=x+1, \;\ du=dx\)

Change the limits of integration.

\(\displaystyle \int_{1}^{3}ln(u)du\)

Now, finish?.

 

[galactus]

As far as using parts to evaluate \(\displaystyle \int\frac{x}{x+1}dx\). I would just rewrite it as:

\(\displaystyle \int dx-\int\frac{1}{x+1}dx\).

Now it is easy to see the solution, x-ln(x+1)