Integration Problem

  • Thread starter derekjn
  • Start date
  • #1
12
0

Homework Statement



[tex]\int\frac{1}{xln(x)^2}[/tex]

Homework Equations



[tex]\frac{d}{dx}[/tex]ln(x) = [tex]\frac{1}{x}[/tex]

The Attempt at a Solution



I made a u substitution, letting u = ln(x); du = [tex]\frac{1}{x}[/tex]dx. The antiderivative I then get is ln(u[tex]^{2}[/tex]) + C but obviously this is not the correct answer. Any help anyone could provide would be greatly appreciated.

Derek
 

Answers and Replies

  • #2
246
6
Have you tried integrating by parts?
 
  • #3
12
0
Thanks for your reply. Unfortunately we haven't covered integration by parts yet; we have only made it as far as logarithmic differentiation so the techniques that I know are limited. derek
 
  • #4
302
1
the procedure you are doing seems correct but your final answer is not correct. You are therefore making an error in the process. Please post your full solution and perhaps someone will point out the error. Or if this helps the answer should be

-1/lnx + c
 
  • #5
1,753
1
Integration by parts is not necessary at all. Just rework your problem with the substitution you chose, and you'll see the error you are making.
 
  • #6
12
0
Thanks for your replies guys. I'm still not seeing it. After the substitution I get [tex]\int\frac{1}{u^{2}}du = ln(u^{2}) + C = 2ln(u) + C[/tex]. I'm stumped at this point.
 
  • #7
185
4
[itex] \int x^{-2} dx = -1/x [/itex].
 
  • #8
12
0
Beautiful. I got so caught up in [tex]\frac{d}{dx}ln(x) = \frac{1}{x}[/tex] that the square didn't even register. Thanks everyone for your help. Derek
 
  • #9
answer comes out to be 1/2(ln(ln(x^2))+c
 

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