Answer :

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]