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Annoying integral

  • Thread starter mugzieee
  • Start date
75
0
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

Science Advisor
Homework Helper
Insights Author
12,949
532
Make a substitution first.

[tex] 2x+1 = u [/tex]

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

Daniel.
 
45
0
Math is easy once you accept the fact that substitution is the way to go ;)
 
75
0
so make the u=2x+1 substitution in the first step of the problm?
 

Pyrrhus

Homework Helper
2,160
1
Yes as dexter noted

[tex] \int \ln (2x+1) dx [/tex]

[tex] u = 2x +1 [/tex]

[tex] du = 2dx [/tex]

[tex] \int \ln (u) \frac{du}{2} [/tex]
 
75
0
gotcha, thanx.
 

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