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Where did this 1/4*ln(C) term come from when integrating?

  1. Aug 26, 2014 #1
    1. The problem statement, all variables and given/known data
    The givenequation is this: [tex]
    \frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}[/tex]
    My book says that when integrated, the above equation becomes [tex]
    \frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c) [/tex]

    I understand everything except for the final term of the last equation. where did the [itex] \frac{1}{4} \ln (c) [/itex] come from and shouldn't it just be c as the constant of integration?
    2. Relevant equations
    [tex]\frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}[/tex]
    [tex]\frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c) [/tex]

    3. The attempt at a solution

    Integrating [tex] \frac{dx}{x} [/tex] gives you [tex]\ln(x) + c[/tex].
  2. jcsd
  3. Aug 26, 2014 #2


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    [itex]\frac14 \ln c [/itex] is an arbitrary constant. Presumably writing it in that form rather than as [itex]D[/itex] makes solving for [itex]u[/itex] much more convenient.

    If [itex]f: U \to \mathbb{R}[/itex] is a surjection, then it is perfectly legitimate to take [itex]f(C)[/itex] for some [itex]C \in U \subset \mathbb{R}[/itex] as a constant of integration rather than [itex]D \in \mathbb{R}[/itex].
  4. Aug 26, 2014 #3
    So then any term as the constant would make sense? E.g. the book could've used [itex] \frac{1}{8} \ln(c) \text{or}\ 3\ln(c) [/itex] as the last term in the integrated equation? I'm just a little confused how they got [itex] \frac{1}{4} \ln(c) [/itex].
  5. Aug 26, 2014 #4


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    Try using a plain old ##c## for the arbitrary constant and solve for ##u##. What happens then if you replace ##c## with ##\frac 14 \ln c##? You should, I hope, see why that form was chosen.
  6. Aug 26, 2014 #5
    I got that [tex] u = \frac{2cx^4-2}{1+cx^4} [/tex]. I still don't see why the form was chosen.
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