annoying integral

[mugzieee]

Hey guys i keep getting stuck with this:
integral of ( ln(2x+1)dx)
im supposed to use by parts
heres what i have done
u=ln(2x+1)
du=2/2x+1
dv=dx
v=x

then i apply the formula uv-vdu
and i end up with another integral im supposed to use by parts for:
integral of(2x/2x+1 dx)
heres what i have done for that integral
u=2x
.5du=dx
dv=1/2x+1
v=ln(2x+1)

then i apply the formula again, and since i have the same integal i stared wit, i add it to the lef side etc etc... but i dont get the correct answer...what does it look lie im doing wrong?

[dextercioby]

Make a substitution first.

2x+1 = u

And then you'd have to integrate something proportional to \int \ln u \ du which is really easy.

Daniel.

[Nerro]

Math is easy once you accept the fact that substitution is the way to go ;)

[mugzieee]

so make the u=2x+1 substitution in the first step of the problm?

[Pyrrhus]

Yes as dexter noted

\int \ln (2x+1) dx

u = 2x +1

du = 2dx

\int \ln (u) \frac{du}{2}

[mugzieee]

gotcha, thanx.