Integrating Natural Log Function using Integration by Parts Method

[jrmed13]

Integrating Natural Log Function using "Integration by Parts" Method

Homework Statement

The problem says to integrate ln(2x+1)dx

Homework Equations

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

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!

 

[rocomath]

\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.

 

[jrmed13]

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

 

[kevinkjt2000]

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