Answer
$$\ln x$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt \cr
& {\text{Integrating }}\int_x^{x + h} {\ln t} dt{\text{ by parts}} \cr
& \int {\ln t} dt,{\text{ }}u = \ln t,{\text{ }}du = \frac{1}{t}dt,{\text{ }}dv = dt,{\text{ }}v = t \cr
& \int {\ln t} dt = t\ln t - \int {t\left( {\frac{1}{t}} \right)dt} = t\ln t - t + C,{\text{ then}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt = \mathop {\lim }\limits_{h \to 0} \left( {\frac{1}{h}\left[ {t\ln t - t} \right]_x^{x + h}} \right) \cr
& {\text{Evaluating limits of integration}} \cr
& = \mathop {\lim }\limits_{h \to 0} \left( {\frac{1}{h}\left[ {\left( {x + h} \right)\ln \left( {x + h} \right) - \left( {x + h} \right)} \right] - \frac{1}{h}\left[ {x\ln x - x} \right]} \right) \cr
& {\text{Distribute}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h} \right)\ln \left( {x + h} \right) - \left( {x + h} \right) - \left( {x\ln x - x} \right)}}{h} \cr
& {\text{By the definiton of the derivative }}f'\left( x \right)\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& {\text{Then,}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt = \frac{d}{{dx}}\left[ {\left( {x\ln x - x} \right)} \right] \cr
& {\text{Differentiating}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt = x\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ x \right] - \frac{d}{{dx}}\left[ x \right] \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt = x\left( {\frac{1}{x}} \right) + \ln x\left[ 1 \right] - 1 \cr
& \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} {\ln t} dt = \ln x \cr} $$
