In the integration formula $\int dx/x = log x + c$, Is the log natural or log base 10? In the integration formula $\int dx/x = log x + c$, Is the log natural or log base 10? The formula appears in many problems and i just got a problem wrong for apparently using the wrong log. Could you please enlighten me about the right log to be used in integration.
 A: Consider the differential equation $y'=y$ with $y(0)=1$ on $\mathbb{R}$. It can be proven that this has a unique solution. We define the exponential function on $\mathbb{R}$ to be equal to this solution. We denote it as $\exp$, but many people abuse notation and denote it $e^x$, which in this context, is fine, as long as everyone knows what is being meant. Now, it can be proven that $\exp$ is function whose range is $\mathbb{R}^+$. In other words, $\exp[\mathbb{R}]=\mathbb{R}^+$. It can also be proven that $\exp$ is a function that has a compositional inverse. This compositional inverse has a name: the natural logarithm, and we denote it $\ln$, or $\log$. Of course, one must remember that $\ln[\mathbb{R}^+]=\mathbb{R}$. The relationship that this function has with other logarithmic functions is complicated, but in some sense, it is the logarithmic function, due to the fact that it can be defined independently of any notion of powers. Hence the name "natural" logarithm. In terms of other logarithmic functions, the natural logarithm can be interpreted as being base $e$, where $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.
Now, given the equation $\exp'=\exp$, it can be proven via the chain rule that $\ln'(x)=\frac{1}{x}$ for every $x\in\mathbb{R}^+$. This is because $(\exp\circ\ln)'(x)=1$, yet by the chain rule, $(\exp\circ\ln)'(x)=(\exp'\circ\ln)(x)\ln'(x)=(\exp\circ\ln)(x)\ln'(x)=x\ln'(x)$, so $x\ln'(x)=1$, implying $\ln'(x)=\frac{1}{x}$. Because of this, $\ln(x)$ is an antiderivative of $\frac{1}{x}$ on $\mathbb{R}^+$.
However, this is not the complete story. Consider the function $\ln^-$ defined by $(\ln^-)(x)=\ln(-x)$ for every $\mathbb{R}^-$. It turns out that $(\ln^-)'(x)=\frac{1}{x}$ is true on $\mathbb{R}^-$ as well! So to talk about the antiderivatives of $\frac{1}{x}$ properly, we need to consider functions $f_{A,B}:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ such that $$f_{A,B}(x)=\begin{cases} \ln(-x)+A & x\lt0 \\ \ln(x)+B & x\gt0 \end{cases}.$$ These functions satisfy the property that $f_{A,B}'(x)=\frac{1}{x}$ for every $x\in\mathbb{R}\setminus\{0\}$. And in fact, this is every function that satisfies the equation on that domain, there are no other functions. So you can say $$\int\,\frac{\mathrm{d}x}{x}=f_{A,B}(x).$$ If you want to limit yourself to $x\gt0$, though, then $$\int\,\frac{\mathrm{d}x}{x}=\ln(x)+B.$$ Hopefully, this covers everything.
A: In the formula
$$
\int \frac{1}{x}\; dx = \log(x)+C
$$
the logarithm is with base $e$. Most calculus books write
$$
\int \frac{1}{x}\; dx = \log|x|+C\tag{1}
$$
with the absolute value on the right-hand side. The formula (1) is understood as $\log|x|$ is an antiderivative of $\frac1x$ on any open interval contained in the set $\mathbb{R}\setminus\{0\}$.
[If one wants to consider any domain for the function $\frac1x$ that is not connected, for instance, its natural domain $(-\infty,0)\cup(0,\infty)$, then one needs more than one "artibrary constants" and different cases for each connected component of the domain on the right-hand side. In other words, the general solutions to the differential equation
$$
\frac{dy}{dx}=\frac{1}{x}
$$
on an open subset that is not connected has more than one "arbitrary constant" C's.]

In many other contexts, the base may not be important due to the fact that they only cause a factor of a constant. But when doing integrals, the natural log is the one that is an antiderivative of $\frac1x$.
A: The formula is $$\int\,\frac{\mathrm{d}x}{x}=\ln(|x|)+C$$
For the definition of the natural logarithm as an integral we have
$$ Ln (x) = \int _1^x \frac {1}{t} dt , x>0 $$
