Consider the following simple integral. \begin{align*} \int\dfrac{1}{2x}\,dx. \end{align*} Now, I would usually work as follows: \begin{align*} \int\dfrac{1}{2x}\,dx=\dfrac{1}{2}\int\dfrac{1}{x}\,dx=\dfrac{1}{2}\ln x+C. \end{align*} However, I thought I would try: \begin{align*} \int\dfrac{1}{2x}\,dx=\dfrac{1}{2}\int\dfrac{2}{2x}\,dx=\dfrac{1}{2}\ln 2x+C. \end{align*} I suppose this is incorrect. Why? Perhaps I should not use the same letter $C$ for both cases?
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$\ln{2x} = \ln{2} + \ln{x}$. You see now why both are correct? – Calculon Apr 17 '15 at 13:41
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$\frac{1}{2}\ln(2x)+C=\frac{1}{2}(\ln(2)+\ln(x))+C=\frac{1}{2}\ln(x)+C_1,$ where $C_1$ is just a different constant: $C_1=C+\frac{1}{2}\ln(2)$. – Nick D. Apr 17 '15 at 13:41
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It's correct because both answers differ by a constant. – Gregory Grant Apr 17 '15 at 13:42
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No, it's not incorrect. Remember that indefinite integral is defined only up to an additive constant, and here in fact
$$\frac12\log 2x +C=\frac12\log x+\frac12\log 2+C$$
So again, the difference is only the addition of a constant, as expected.
Timbuc
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