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Simple integral, textbook seems wrong

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the curve for which the slope is (ln x)^2/x and passes through P(1, 2)

    2. Relevant equations

    3. The attempt at a solution

    Integrate (ln x)^2 = 1/2 Integral( ((ln x)^2) 2/x dx)

    I get: 1/2 [((ln x)^3/3) + C]

    Then solving for C, I get C=2

    Then my final answer is (ln x)^3/6 + 2

    The textbook says it is (ln x)^3/3 + 2.

    I don't get it.
  2. jcsd
  3. Dec 6, 2008 #2
    How exactly did you make this integration: Integral( ((ln x)^2) 2/x dx)
    you should get out [tex]\frac{2ln(x)^3}{3}[/tex]
    Why did you actually put this factor 2 in the integral and divide by 2 again, I don't get why this makes sense.
  4. Dec 7, 2008 #3


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    Homework Helper

    Well I got [tex]\int \frac{(ln x)^2}{x} dx = \frac{1}{3} (ln x)^3 + C [/tex]. Don't see where you got that factor of 1/2 from. C = 2, so that's right.
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