Integral Trouble

  • Thread starter laker88116
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  • #1
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[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.
 

Answers and Replies

  • #2
Galileo
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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.
 
  • #3
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[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]
 
  • #4
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Wow, how did I miss that, thanks so much.
 
  • #5
Hurkyl
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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.
 
  • #6
Hurkyl
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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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