# Evaluating a Definite Integral

1. Jul 12, 2012

### Saterial

1. The problem statement, all variables and given/known data
∫ln(x+1)/(x+1)2dx from a=1, b=2

2. Relevant equations
∫fdg=fg-∫gdf

3. The attempt at a solution
Let u = (x+1), du=dx

∫lnu/u2du
Let f=lnu , df = 1/udu, g=-1/u, dg=1/u2du

=-lnu/u+∫1/u2du
=[-lnu+1/u] 1->2

Using Fundamental Theorem of Calculus plug in a,b

=[-ln(2+1)/(2+1)-(-ln(1+1)/(1+1)]
=-ln3/3+ln2/2

What did I miss in order for me not to get 1/6 in the answer? Help would be great!

Thanks

2. Jul 12, 2012

### CAF123

When you used a change of variable right at the beginning of u= x+1, you have to evaluate your limits of integration for the new variable. Ie after using parts, you should integrate between 3 and 2.

3. Jul 12, 2012

### Staff: Mentor

No, you don't have to evaluate at the new limits when you do a change of variable. If you "undo" the substitution, you can use the original limits.

∫lnu/u2du
Let f=lnu , df = 1/udu, g=-1/u, dg=1/u2du

=-lnu/u+∫1/u2du
=[-lnu+1/u]
You are missing a u in the denominator of the first term, so it should be -ln(u)/u.

Undo the substitution to get -ln(x + 1)/(x+1) + 1/(x + 1) and evaluate at 2 and 1, as usual, and you'll get the answer you showed.

4. Jul 12, 2012

### Saterial

Thanks for the replies!

I tried out both methods but did not have any luck. With the first method provided by CAF123. I was unable to get an answer close to what it should be. With Mark's method. I received an answer of -ln3/3+ln2/2+2. I still cannot find where I am going wrong to not get 1/6.

5. Jul 13, 2012

### SammyS

Staff Emeritus
What is $\displaystyle \int \frac{1}{u^2}\,du\ ?$

(You have a sign error.)

Last edited: Jul 13, 2012
6. Jul 13, 2012

### CAF123

In the end, you should be integrating -ln(u)/u -1/u with upper limit 3, lower limit 2.
Or as Mark44 said, integrate -ln(x+1)/(x+1) -1/(x+1) with upper limit 2, lower limit 1.
(To Mark44: I understood this perfectly, I think I just misinterpreted what the last line was in the OP's attempt).
Many thanks.

7. Jul 13, 2012

### Saterial

How come? isn't that suppose to be ∫gdf? In which, pulling out the negative will make the integral become positive and 1/u^2du.

8. Jul 13, 2012

### CAF123

Correct, but the integral of 1/u^2 is -1/u