Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\ln x} dx \cr
& {\text{Integrate }}\int {\ln xdx{\text{ by parts}}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = dx,{\text{ }}v = x \cr
& \int {\ln x} dx = x\ln x - \int {x\left( {\frac{1}{x}} \right)} dx \cr
& \int {\ln x} dx = x\ln x - \int {dx} \cr
& \int {\ln x} dx = x\ln x - x + C \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\ln x} dx = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {\int_b^1 {\ln x} dx} \right] \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {x\ln x - x} \right]_b^1 \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {1\ln 1 - 1 - \left( {b\ln b - b} \right)} \right] \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \left[ { - 1 - b\ln b + b} \right] \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{b \to {0^ + }} b\ln b + \mathop {\lim }\limits_{b \to {0^ + }} b \cr
& {\text{Where, }}\mathop {\lim }\limits_{b \to {0^ + }} b\ln b = \mathop {\lim }\limits_{b \to {0^ + }} \frac{{\ln b}}{{1/b}} = \frac{\infty }{\infty } \cr
& {\text{By the LHopital's rule}} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \frac{{\ln b}}{{1/b}} = \mathop {\lim }\limits_{b \to {0^ + }} \frac{{1/b}}{{ - 1/{b^2}}} = - \mathop {\lim }\limits_{b \to {0^ + }} b = 0 \cr
& {\text{Then}} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{b \to {0^ + }} b\ln b + \mathop {\lim }\limits_{b \to {0^ + }} b \cr
& {\text{ }} = - 1 - 0 + 0 \cr
& {\text{ }} = - 1 \cr
& \int_0^1 {\ln x} dx = - 1 \cr} $$
