Intergration using logarithms

  • Thread starter kuahji
  • Start date
  • #1
394
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Evaluate the integral.

[tex]\int3/(3x-2) dx[/tex] from 0 to -1 (top to bottom).

I change the equation to [tex(1/x - 3/2) dx[/tex]
then integrated ln x-3/2x, but ln x at 0 is undefined.

The text book shows it as becoming ln (3x-2), but I'm not completely understanding how to get to that.
 

Answers and Replies

  • #2
olgranpappy
Homework Helper
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substitute y = 3x-2
 
  • #3
394
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Thanks, that worked. I'm just a bit rusty I reckon.
 
  • #4
HallsofIvy
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Caution: you are going to find it very difficult to do Calculus if you cannot do basic algebra.

[itex]a/(b+c)[/itex] is not equal to (a/b)+ (a/c)!

You had better review your algebra.
 
  • #5
394
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Caution: you are going to find it very difficult to do Calculus if you cannot do basic algebra.

[itex]a/(b+c)[/itex] is not equal to (a/b)+ (a/c)!
Hmm, guess you're right. I've always used (a+b)/c=a/c+b/c but guess I just assumed it would work vice versa. Is there any combination it does equal?
 
  • #6
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My 1st help in answering !!

1- You have to use change of variables method.
2- Substitute den with a variable and take derivative of this wrt x
3- Using this var the limits will change
4- Once you do 2 and with new limits from 3, the integral will be easy.

Thanks

Asif
 

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