# Integrating Natural Log Function using Integration by Parts Method

1. Oct 4, 2008

### jrmed13

Integrating Natural Log Function using "Integration by Parts" Method

1. The problem statement, all variables and given/known data
The problem says to integrate ln(2x+1)dx

2. Relevant equations
I used u=ln(2x+1); du = 2dx/(2x+1); dv=dx; v=x

3. The attempt at a solution
So, I integrated it using that (above) 'dictionary' and I got the expression xln(2x+1) - integral of (2x/2x+1)

I could substitute again and say that u=(1/(2x+1)); dv=2xdx, but that process would never end!!
And, if I use u=(2x), then various parts of the equation would cancel and I would be left with integral of (ln(2x+1)) = integral of (ln(2x+1))...
I know that the answer should be 0.5(2x+1)ln(2x+1) - x +C, but I can't seem to get it!!

2. Oct 4, 2008

### rocomath

Re: Integrating Natural Log Function using "Integration by Parts" Method

$$\int\ln{(2x+1)}dx=x\ln{(2x+1)}-\int\frac{2x}{2x+1}dx$$

If you add 1 and subtract 1, you can attain your denominator.

$$\int\frac{2x+1-1}{2x+1}dx$$

Now break it up, and go from there.

3. Oct 4, 2008

### jrmed13

Re: Integrating Natural Log Function using "Integration by Parts" Method

THANK YOU SO MUCH!!! (I got the answer :D)

4. Sep 2, 2009

### kevinkjt2000

Re: Integrating Natural Log Function using "Integration by Parts" Method

This is the general rule that can be proved for any type of antideravitve of the the natural log function by use of integration by parts.
$\int ln{(u)} du = u * (-1 + ln{(u)}) + C$