Sorry to bring this thread back up, but I figured this would be better than starting a new one. I have run into another brick wall that is driving me insane (see following picture:

)

It seems simple, yet I'm not getting the answer in the book, but rather I'm looping through the Integration By Parts method over and over again.

$\displaystyle \int{\ln{(2x+1)}dx}$

I set:

$\displaystyle u=\ln{(2x+1)}$ $\displaystyle dv=dx$

$\displaystyle du=\frac{2}{2x+1}$ $\displaystyle v=x$

and get:

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

and from there I integrate by partial fractions:

$\displaystyle 2\int{x\frac{1}{2x+1}}$

$\displaystyle u=\frac{1}{2x+1}$ $\displaystyle dv=x$

$\displaystyle du=\frac{2}{(2x+1)^2}$ $\displaystyle v=\frac{x^2}{2}$

then I get:

$\displaystyle 2[\frac{x^2}{2(2x+1)}-\int{\frac{x^2}{(2x+1)^2}}]$

I might be wrong, but it seems like and endless loop to me

Could anyone help me out?