How to determine the base of the logarithm that results from the evaluation of $\int \frac1x dx$? All textbooks I've used claim that this integral $\int \frac1x dx$ is equal to the natural logarithm, but this is obviously just a special case, as the antiderivative could actually be any logarithm.
Proof
Using derivative definition it is easy to prove that
$\left(  \log_{b}(x) \right)' = \lim _{h\to 0}\left[\frac{\log_{b}\left(x+h\right)− \log_{b}\left(x\right)}{h}\right] \stackrel{\text{De L'Hospital}}{=} \lim _{h\to \:0}\left(\frac{\frac{1}{h+x}}{1}\right) = \frac1x$
Hence, by the fundumental theorem of calculus we know that:
$\int \frac1x dx = \log_b(x)$

Given all that, I don't quite get how to actually determine which logarithm to choose when evaluating the integral.
For example $\int \frac1x dx = \log_{10}(x)$ and $\int \frac1x dx = \log_b(x) = \ln(x)$ but obviously $\log_{10}(x) \neq \ln(x)$
Could someone explain why do I get different results when evaluating the same integral?
 A: Antiderivatives differ by additive constants; logarithms with different bases differ by multiplicative constants. Therefore, only one base is correct. It is, of course, the solution to $\int_1^b\frac{dx}{x}=1$. One way to see $b=e$ is to note $z(y):=\int_1^y\frac{dx}{x}$ solves$$\frac{dz}{dy}=\frac1y\implies\frac{dy}{dz}=y\implies y\propto e^z.$$For what it's worth, your use of L'Hôpital's rule should be$$\log_b^\prime x=\lim_{h\to0}\tfrac1h\log_b(1+h/x)=\frac{1}{\ln b}\underbrace{\lim_{h\to0}\tfrac1h\ln(1+h/x)}_{\frac1x}.$$
A: The answer to your question depends on how you define the natural logarithm; all of these definitions are equivalent and can be proved from each other.  Some authors take $$\int_{t=1}^x \frac{1}{t} \, dt = \ln x,$$ for example.  Others define it in terms of the inverse function of $e^x$ (which in itself has various equivalent definitions for $e$).  Refer to https://en.wikipedia.org/wiki/Natural_logarithm for more information.
A: Using well-known properties of logarithms, for a fixed base $b$ (different from the base of the natural log) and variable $x$ we know that
$$  \log_b (x) = \frac1{\ln(b)} \ln(x). $$
Since the factor $1/\ln(b)$ is a constant, we can apply the usual rule for
the derivative of a constant multiple of a function,
$$
(\log_b (x))' = \frac{d}{dx}  \log_b (x)
 = \frac1{\ln(b)}\left( \frac{d}{dx} \ln(x)\right)
 = \frac1{\ln(b)}\left( \frac1x \right) \neq \frac1x,
$$
and your conclusion that $(\log_b (x))' = 1/x$ is simply incorrect.
The step labeled "L'Hospital" seems to be the culprit.
Applying the fundamental theorem then gives us
$$ \frac1{\ln(b)}\int \frac1x \,dx
= \int \frac1{\ln(b)}\left( \frac1x \right)\,dx
= \log_b(x) + C,
$$
which is just another way of saying that
$$ \int \frac1x \,dx
= \ln(b) \log_b(x) + C_2 = \ln(x) + C_2,
$$
which of course we already knew.
