Answer:
[tex]\dfrac{1}{3}\ln|3x-8|+C[/tex]
Step-by-step explanation:
Given indefinite integral:
[tex]\displaystyle \int \dfrac{1}{3x-8}\;dx[/tex]
To evaluate the indefinite integral, we can use the method of substitution.
[tex]\textsf{Let}\;\;u=3x-8[/tex]
Find du/dx and rearrange to isolate dx:
[tex]\dfrac{du}{dx}=3 \implies dx=\dfrac{1}{3}\;du[/tex]
Rewrite the original integral in terms of u and du:
[tex]\displaystyle \int \dfrac{1}{u}\cdot \dfrac{1}{3}\;du[/tex]
Take out the constant:
[tex]\displaystyle \dfrac{1}{3} \int \dfrac{1}{u}\;du[/tex]
Now, evaluate using the common integral ∫ 1/x dx = ln|x|:
[tex]\dfrac{1}{3}\ln|u|+C[/tex]
Substitute back in u = 3x + 8:
[tex]\dfrac{1}{3}\ln|3x-8|+C[/tex]
Therefore:
[tex]\displaystyle \int \dfrac{1}{3x-8}\;dx=\boxed{\dfrac{1}{3}\ln|3x-8|+C}[/tex]
