Answer
$2x\displaystyle \ln x+\frac{x^{2}+1}{x}$
Work Step by Step
$u(x)=x^{2}+1,\qquad u^{\prime}(x)=2x$
$v(x)=\displaystyle \ln x,\qquad v^{\prime}(x)=\frac{1}{x}$
$f(x)=u(x)v(x)$, so by the Product Rule,
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=2x\displaystyle \cdot\ln x+(x^{2}+1)\cdot\frac{1}{x}$
$= 2x\displaystyle \ln x+\frac{x^{2}+1}{x}$
