Evaluating the Integral of 1/x

Since we mentioned earlier that integrals are the reverse of derivatives, then we let $f(x)=1/x$. So that we have:
\begin{align*}
\int\dfrac{1}{x}\,dx=F(x)+C,
\end{align*}

where:
\begin{align*}
\dfrac{d}{dx} F(x)=\dfrac{1}{x}.
\end{align*}

Note that the derivative of $\ln ⁡x$ is $1/x$. Thus, it follows that:
\begin{align*}
\dfrac{d}{dx} \ln⁡ x=\dfrac{1}{x},
\end{align*}

then:
\begin{align*}
\int\dfrac{1}{x}\, dx=\ln⁡ x+C.
\end{align*}

However, we will notice that the only restrictions in the domain of $f’(x)$, which is $x$, must not be equal to $0$. So, in $f’(x)$, $x>0$ or $x<0$, but $x\neq0$. While in the function $\ln ⁡x$, the domain is only the positive numbers since natural logarithmic is not defined in negative numbers or in $0$. Hence, $x$ is strictly a positive number.

This follows that $1/x$ and $\ln⁡(x)$ have different domains, which is not okay since they must have the same domain. So we need to consider when $x<0$.

To do this, we need to assume that $x=-u$, where $u$ is a real number. This follows that if $x<0$, then $u>0$. And substituting the value of $x$, we will have $dx=-du$, and this implies that:
\begin{align*}
\int\left(\dfrac{1}{x}\right)\, dx=\int\left(\dfrac{1}{-u}\right)\,\left(-du\right).
\end{align*}

This follows that when $x<0$, then the integral of $f'(x)$ is:
\begin{align*}
\int\left(\dfrac{1}{x}\right)\, dx= \ln (u)+C_1,
\end{align*}

where $C_1$ is an arbitrary constant. And by substituting the value of $u$, we have:
\begin{align*}
\int\left(\dfrac{1}{x}\right)\, dx= \ln (-x)+C_1.
\end{align*}

However, we know that the natural logarithmic is not defined in negative numbers, hence we will use the absolute function, where if $x\geq0$, then $|x|=x$, and if $x<0$, then $|x|=-x$. Therefore, the integral of $1/x$ is $\ln⁡|x|+C$, where $C$ is an arbitrary constant.

Thus, this verifies and explains the integral of $1/x$ proof.