Lillianna Andersen

2022-07-03

Two apparently different antiderivatives of $\frac{1}{2x}$
What is right way to calculate this integral and why?
$\int \frac{1}{2x}\text{d}x$
I thought, that this substitution is right:
$t=2x$
$\text{d}t=2\text{d}x$
$\frac{\text{d}t}{2}=\text{d}x$
$\int \frac{1}{2x}\text{d}x=\int \frac{1}{t}\frac{\text{d}t}{2}=\frac{1}{2}\mathrm{ln}|2x|+C.$
But it's not right, because this is the correct answer:
$\int \frac{1}{2x}\text{d}x=\frac{1}{2}\int \frac{1}{x}\text{d}x=\frac{1}{2}\mathrm{ln}|x|+C.$
Can someone explain me, why is the first way wrong?

Step 1
The two answers actually agree, that is, the first calculation really does produce an antiderivative of $\frac{1}{2x}$ as expected, at least provided we interpret C to be a general constant in both cases. Expanding the first antiderivative gives $\frac{1}{2}\mathrm{ln}|2x|+C=\frac{1}{2}\left(\mathrm{log}|x|+\mathrm{log}2\right)+C=\frac{1}{2}\mathrm{log}|x|+\left(C+\frac{1}{2}\mathrm{log}2\right).$
Step 2
So, if we denote ${C}^{\prime }:=C+\frac{1}{2}\mathrm{log}2$, this antiderivative is $\frac{1}{2}\mathrm{ln}|x|+{C}^{\prime },$, which (again, regarded as a family of functions, all equal up to an over constant) coincides with the second antiderivative.
This exemplifies the statement that an antiderivative of a given function is unique up to addition of an overall constant. (Actually, this only need be true when the domain of the function is connected, which in particular is not the case for the given integrand, but this isn't essential to the question at hand.)

Sylvia Byrd

Explanation:
You are correct.
Since $\mathrm{ln}|2x|=\mathrm{ln}|x|+\mathrm{ln}2$

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