Answer:
[tex]\displaystyle \int \dfrac{1}{x}\ dx = \ln|x| + C[/tex]
Step-by-step explanation:
We are evaluating the indefinite integral:
[tex]\displaystyle \int \dfrac{1}{x}\ dx[/tex]
First, we can make the substitution:
[tex]\text{Let } x = e^t \ \ \ \text{Then:}[/tex]
[tex]dx = e^t\ dt[/tex]
The integral becomes:
[tex]\displaystyle \int \dfrac{1}{e^t}\!\left(e^t\ dt\right)[/tex]
[tex]\displaystyle = \int 1\ dt[/tex]
Applying the power rule, we get:
[tex]\displaystyle \int 1\ dt = t + C[/tex]
Then, we can plug back in the definition for t in terms of x:
[tex]x = e^t \ \ \implies \ \ t = \ln|x|[/tex]
Thus, the integral evaluates to:
[tex]\huge\boxed{\int \dfrac{1}{x}\ dx = \ln|x| + C}[/tex]
