integration by parts - using a previous integration

Hello

I am working on a problem which involves integrating [with limits 0, 1 (don't know how to do limits in latex)].

So I managed to integrate using substitution as follows:

let u = 1+x, du/dx = 1, limits: u = 1+1 = 2 and u = 1+0=1

x = u-1.

So get:

[with different limits 1, 2]

I expand out:

and then:

and I get a result of 5/6 - ln2.

I am then asked to integrate this: with limits 0 to 1.

So I start integrating (by parts) and notice that second integration bit - I have this:

=

That last bit, I already integrated above.

So my question is, how do I combine the two. My start part of integration by parts and then the minus - and here use my previous result.

The previous result had different limits - so I presume I cant just use the result. Do I just substitute back in 1+x for the u above like this:

- then use my limits 0, 1 as asked for this integration?

Angus

Re: integration by parts - using a previous integration

You can say that

Re: integration by parts - using a previous integration

What you got is this: ?

You already know the value of , just divide it by 3 and subtract, yielding: