Solving Integral of ln(2x+1)dx with By Parts Method

[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 I am 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 don't get the correct answer...what does it look lie I am 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.