Answer
$$\overline f \left( x \right) = \frac{1}{2}\left[ {x\ln x - x\ln x\left( {x - 2} \right) + 2\ln \left( {x - 2} \right) - 2} \right]$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln x \cr
& {\text{The }}n{\text{ - unit moving average of a function }}f{\text{ is}} \cr
& \overline f \left( x \right) = \frac{1}{n}\int_{x - n}^x {f\left( t \right)} dt \cr
& {\text{The 2 - unit moving average of }}f\left( x \right) = \ln x{\text{ is}} \cr
& \overline f \left( x \right) = \frac{1}{2}\int_{x - 2}^x {\ln t} dt \cr
& {\text{Integrating}} \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {t\ln t - t} \right]_{x - 2}^x \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\left( {x\ln x - x} \right) - \left( {x - 2} \right)\ln \left( {x - 2} \right) + \left( {x - 2} \right)} \right] \cr
& {\text{Simplifying}} \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {x\ln x - x - x\ln x\left( {x - 2} \right) + 2\ln \left( {x - 2} \right) + x - 2} \right] \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {x\ln x - x\ln x\left( {x - 2} \right) + 2\ln \left( {x - 2} \right) - 2} \right] \cr} $$
