#### Answer

$(f+g)(x)=\dfrac{3x^2-x+1}{x}$

#### Work Step by Step

Since $(f+g)(x)=f(x)+g(x),$ with $f(x)=3x-1$ and $g(x)=\dfrac{1}{x},$ then
\begin{array}{l}\require{cancel}
(f+g)(x)=f(x)+g(x)
\\\\
(f+g)(x)=3x-1+\dfrac{1}{x}
\\\\
(f+g)(x)=(3x-1)\cdot\dfrac{x}{x}+\dfrac{1}{x}
\\\\
(f+g)(x)=\dfrac{3x^2-x}{x}+\dfrac{1}{x}
\\\\
(f+g)(x)=\dfrac{3x^2-x+1}{x}
.\end{array}