Answer
$(\dfrac{x^2}{2} \log_2 x-\dfrac{x^2}{4 \ln 2})+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int x \log_2 x \ dx=\int \dfrac{x \ln x}{\ln 2} \ dx$
or, $=\dfrac{1}{\ln 2} \ln x(\dfrac{x^2}{2})-\dfrac{1}{\ln 2} \int (\dfrac{x^2}{2}) \dfrac{1}{x} \ dx$
or, $=\dfrac{x^2 \ln x}{2 \ln 2}-\dfrac{1}{2 \ln 2} (\dfrac{x^2}{2})+C$
or, $=(\dfrac{x^2}{2} \log_2 x-\dfrac{x^2}{4 \ln 2})+C$
