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Integration problems

  1. Jul 24, 2014 #1
    Hey!

    I have this integral: ∫((1/2)/(2x-1))dx.

    The first time, I did like this: ∫((1/2)/(2x-1))dx = (1/2)∫(1/(2x-1))dx. If I set u = 2x-1, then du = 2dx, so I can rewrite (1/2)∫(1/(2x-1))dx as (1/2)*(1/2)∫(1/u)du = 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|2x-1|.

    But when I do like this (I cannot see what im doing wrong) ∫((1/2)/(2x-1))dx = ∫(1/(4x-2)dx and set u = 4x-2, so du = 4dx, I can rewrite it as 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|4x-2|. What am I doing wrong??

    Please help me out! Thanks!
     
  2. jcsd
  3. Jul 24, 2014 #2
    The error you made is that you forgot the constant of integration. Include that, and recall that [itex]\ln{ab} = \ln{a} + \ln{b}[/itex] for positive, non-zero a and b, you should be able to see that the methods are equivalent.
     
  4. Jul 24, 2014 #3

    jbunniii

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    Note that
    $$\begin{align}
    \frac{1}{4}\ln|4x-2| &= \frac{1}{4}\ln(2|2x-1|) \\
    &= \frac{1}{4}(\ln(2) + \ln|2x-1|)\\
    &= \frac{1}{4}\ln|2x-1| + \frac{\ln(2)}{4}
    \end{align}$$
    which differs from your other answer ##\frac{1}{4}|2x-1|## by a constant. Since indefinite integrals can differ by a constant (##+C##), both answers are equally valid.
     
  5. Jul 24, 2014 #4
    Thanks a lot, i understand now!
     
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