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Math Help - just a test

  1. #1
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    just a test

    <br />
\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
    Last edited by harold; October 4th 2008 at 11:10 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by harold View Post
    <br />
\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!
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  3. #3
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    Thanks Jhevon!! Was wondering why it didn't have the look it should. Good catch with the absolute value signs too!
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