Answer
$\ln (2x) (\dfrac{x^3}{3})-\dfrac{x^3}{9}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln (2x)$ and $dv=x dx \implies v=\dfrac{x^3}{3}$
$\displaystyle \int x^2 \ln (2x) \ dx=\ln (2x) (\dfrac{x^3}{3})-\int (\dfrac{x^3}{3}) (\dfrac{1}{x}) \ dx$
or, $=\ln (2x) (\dfrac{x^3}{3})-\dfrac{1}{3} \int x^2 \ dx$
or, $= \ln (2x) (\dfrac{x^3}{3})-\dfrac{x^3}{9}+C$
