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

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

$\Rightarrow$ $\log x \cdot \int {\left( {{x^2} + 1} \right)dx} - \int {\left( {\cfrac{d}{{dx}}\left( {\log x} \right) \cdot \int {\left( {{x^2} + 1} \right)dx} } \right)dx}$

$= \log x \cdot \left( {\cfrac{{{x^3}}}{3} + x} \right) - \int {\cfrac{1}{x}\left( {\cfrac{{{x^3}}}{3} + x} \right)} dx + C$

$= \left( {\cfrac{{{x^3}}}{3} + x} \right)\log x - \int {\left( {\cfrac{{{x^2}}}{3} + 1} \right)} dx + C$

$= \left( {\cfrac{{{x^3}}}{3} + x} \right)\log x - \cfrac{{{x^3}}}{9} - x + C$