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Integral Trouble

  • Thread starter laker88116
  • Start date
57
0
[tex]\int ln(2x+1)dx[/tex]

So far I know that I need to use integration by parts, I let [itex]u= ln(2x+1) [/itex] and so [itex] du= \frac {dx}{2x+1} [/itex]. Also, I said [itex] dv= dx [/itex] and [itex] v=x [/itex].

So then plugging this into the equation for integration I get:

[tex] xln(2x+1) - \int \frac {2x}{2x+1}dx [/tex]

Then I determine that I need to do integration by parts again on the latter half of the function. So, for [itex]\int \frac {2x}{2x+1}dx [/itex], I let [itex]u= 2x [/itex] and so [itex] du= xdx [/itex]. Also, I said [itex] dv= \frac {dx}{2x+1} [/itex] and [itex] v= \frac {ln(2x+1)}{2} [/itex].

So then plugging this into the equation for integration I get:

[tex]\int \frac {2x}{2x+1}dx = xln(2x+1) - \int ln(2x+1)dx [/tex]

Now, I have like terms so I say that:

[tex]\int ln(2x+1)dx = xln(2x+1) - [xln(2x+1) - \int ln(2x+1)dx]. [/tex]

I am not sure where I made an error here. Any help is appreciated.
 

Galileo

Science Advisor
Homework Helper
1,989
6
Oi, so complicated >_<

You don't have to integrate by parts again. You can split the fraction like:

[tex]\frac{2x}{2x+1}=1-\frac{1}{2x+1}[/tex]

But even simpler is first solving:

[tex]\int \ln x dx[/tex]
then the integral should be a piece of cake. The straight way to Rome is not always the shortest, nor the easiest to follow.
 
128
0
[tex]\int ln(2x+1)dx = xln(2x+1) - \int \frac {2x}{2x+1}dx = xln(2x+1) - \int 1 - \frac {1}{2x+1} dx[/tex]
 
57
0
Wow, how did I miss that, thanks so much.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
Anyways, as to the actual method you implemented, your second integration was simply undoing the first integration by parts you tried. That's why you got a useless result at the end.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
It's not productive to do the problem for the person asking for help. (especially when you do it wrong!) Fortunately, the poster had already figured it out from the hints!
 

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