$\cfrac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}$

: Let $I = \int {\cfrac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}} dx = \int {{{\left( {x + \log x} \right)}^2}\left( {1 + \cfrac{1}{x}} \right)dx}$

Put $x + \log x = t$ $\Rightarrow$ $\left( {1 + \cfrac{1}{x}} \right)dx = dt$

$\therefore$ $I = \int {{t^2}} dt = \cfrac{{{t^3}}}{3} + C = \cfrac{1}{3}{\left( {x + \log x} \right)^3} + C$