text: # just a test
  
  • Oct 3rd 2008, 01:19 PM
  harold
  just a test
  $
  \int \frac {dx}{x \ln x^2} dx$
  
  1) $= \int \frac {dx}{2x \ln x} dx$ This step follows from the property of logs (exponents)
  
  2) $= \frac {1}{2}\int \frac {dx}{x \ln x} dx$ Extract any constants and place in front of the integral
  
  Let $u = \ln x$, then $du = \frac {1}{x} dx.$
  
  3) $= \frac {1}{2}\int \frac {1}{\ln x} \cdot \bigg (\frac{1}{x}\bigg)dx$
  
  4) $= \frac {1}{2}\int \frac {1}{u} du$ U-substitution applied
  
  5) $= \frac {1}{2} \ln {\color{black}|}u{\color{black}|} + C$ By Formula C, and $u > 0$
  
  6) $= \frac {1}{2} \ln {\color{black}|}(\ln x){\color{black}|} + C$ By Reverse substitution
  • Oct 3rd 2008, 02:04 PM
  Jhevon
  Quote:
  
  Originally Posted by harold
  $
  \int \frac {dx}{x ln x^2} dx$
  
  1) $= \int \frac {dx}{2x ln x} dx$ This step follows from the property of logs (exponents)
  
  2) $= \frac {1}{2}\int \frac {dx}{x ln x} dx$ Extract any constants and place in front of the integral
  
  Let $u = ln x$, then $du = \frac {1}{x} dx.$
  
  3) $= \frac {1}{2}\int \frac {1}{ln x} \cdot \bigg (\frac{1}{x}\bigg)dx$
  
  4) $= \frac {1}{2}\int \frac {1}{u} du$ U-substitution applied
  
  5) $= \frac {1}{2} ln u + C$ By Formula C, and $u > 0$
  
  6) $= \frac {1}{2} ln (ln x) + C$ By Reverse substitution
  
  i know this is just a LaTeX test, but just to be a pain.
  
  type \ln x, instead of just ln x, it looks nicer. you get $\ln x$ as opposed to $ln x$, see?
  
  also, $\ln (x^2) = 2 \ln {\color{red}|}x {\color{red}|}$ ....yes, the absolute valus matter
  
  and you also should have absolute values in your solutions, so it is $\frac 12 \ln {\color{red}|} \ln x {\color{red}|} + C$
  
  other than that, good job!
  • Oct 4th 2008, 11:13 AM
  harold
  Thanks Jhevon!! Was wondering why it didn't have the look it should. Good catch with the absolute value signs too!