Answer
$y=\ln |x|+3x+C$
Work Step by Step
We are given that $\dfrac{dy}{dx}=\dfrac{1}{x}+3$
We will separate the variables to obtain:
$dy=(\dfrac{1}{x}+3) dx$
Integrate to obtain:
$\int dy=\int(\dfrac{1}{x}+3) dx$
This implies that $y=\ln |x|+3x+C$
