Answer
\[ = \frac{{{x^2} + 1}}{x} + 2x\ln x\]
Work Step by Step
\[\begin{gathered}
\frac{d}{{dx}}\,\,\left[ {\,\left( {{x^2} + 1} \right)\ln x} \right] \hfill \\
\hfill \\
Use\,\,product\,\,rule \hfill \\
\hfill \\
= \,\left( {{x^2} + 1} \right)\frac{d}{{dx}}\,\,\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\,\left( {{x^2} + 1} \right) \hfill \\
\hfill \\
Differentiate\,\,use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
= \,\left( {{x^2} + 1} \right)\,\left( {\frac{1}{x}} \right) + \ln x\,\left( {2x} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{{x^2} + 1}}{x} + 2x\ln x \hfill \\
\end{gathered} \]
