Answer
$(x \log_3 x-\dfrac{x}{\ln 3})+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int \log_3 x \ dx=\int \dfrac{\ln x}{\ln 3} \ dx$
or, $=\dfrac{1}{\ln 3} \ln x(x)-\dfrac{1}{\ln 3} \int x \times \dfrac{1}{x} \ dx$
or, $=\dfrac{x \ln x}{\ln 3}-\dfrac{1}{\ln 3} x+C$
or, $=(x \log_3 x-\dfrac{x}{\ln 3})+C$
