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When you calculate $\int\frac{1}{2x}dx$ you get $\frac{1}{2}\ln(x)$ and when you calculate $\int\frac{1}{2x}dx$ you don't get $\frac{1}{2}\ln(2x)$.

$\frac{1}{2x}$ is the same as $\frac{1}{2x}$

why do you get different answers?

kjetil b halvorsen
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Yusaf
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    " $\frac{1}{2x}$ is the same as $\frac{1}{2x}$ ". And $ \frac{1}{2} \ln (x) \ $ is the same as $ \frac{1}{2} \ln (2x) \ , $ with a "vertical shift" when you graph the two functions. When we calculate indefinite integrals, we are not finding a single function, but an entire "family" of functions which differ only by a constant; so, they all have the same derivative function. That is the meaning of the arbitrary constant, as the posters are describing in their answers. – colormegone Mar 28 '14 at 13:18

2 Answers2

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$$\frac{1}{2}\ln(2x) = \frac{1}{2}(\ln 2 + \ln x) = \frac{\ln2}{2} + \frac12\ln x$$

When you integrate $$\frac{\frac12}{x},$$ you get $\frac12 \ln x + C$ ($C$ is the constant you are always nagged about!), not just $\frac12\ln x$

5xum
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    +1. Never forget the constant of integration when doing indefinite integrals. – Arthur Mar 28 '14 at 12:59
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    Questions like this are very good in my oppinion, as they really show that the $+C$ is not just for show. – 5xum Mar 28 '14 at 12:59
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I think that @5xum has already said it, but to make it super clear:

The integral of $1 / (2x)$ is not equal to $\frac{1}{2}\ln(x)$ or $\frac{1}{2}\ln(2x)$. you have $$ \int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C. $$ As noted in the other answer and in the comments, this $+C$ isn't just a pretty thing. It is very important because without it, the answer is wrong. The example that you have given shows exactly why it is important. So you have, for example, $$ \int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C_1 = \frac{1}{2}\ln(2x) + C_2 $$ where $C_1$ and $C_2$ are different constants. The reason that this is true is because of the definition. Remember that we say that $$ \int f(x)\; dx = F(x) + C $$ exactly when $F'(x) = f(x)$. And indeed you have that $$\begin{align*} \frac{d}{dx} \frac{1}{2}\ln(x) = \frac{d}{dx} \frac{1}{2}\ln(2x) = \frac{1}{2x}. \end{align*} $$

Thomas
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  • It's not so much that the answer is "wrong", as that the result is incomplete. An indefinite integral produces an infinite set of anti-derivative functions, all having the same derivative function, which is the integrand. – colormegone Mar 28 '14 at 13:24
  • already accepted answer but this is great :) +1 – Yusaf Mar 28 '14 at 13:34