Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\ln x} dx \cr
& {\text{The domain of the function }}f\left( x \right) = \ln x{\text{ is }}\left( {0,\infty } \right),\cr
& {\text{so the integral is not defined for x=0}} \cr
& {\text{By the definition of the improper integrals, we have}} \cr
& \int_0^1 {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx \cr
& \cr
& {\text{Integrate }}\int {\ln x} dx,{\text{ Using the method of integration by parts}} \cr
& \,\,\,\,\,\,\,\,\,{\text{Let }}u = \ln x,\,\,\,\,\,du = \frac{1}{x}dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx,\,\,\,\,\,v = x \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \to \,\,\,\,\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
& \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \left( {x\ln x - x} \right)_a^1 \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( {1\ln 1 - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( { - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \left( {a\ln a} \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr
& \cr
& {\text{Write }}\mathop {\lim }\limits_{a \to {0^ + }} \left( {a\ln a} \right){\text{ as }}\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr
& {\text{Using the L'Hopital's rule for }}\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \frac{{\left( {1/a} \right)}}{{ - \left( {1/{a^2}} \right)}} + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) + 2\mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr
& {\text{Evaluate the limit when }}a \to {0^ + } \cr
& = \left( { - 1} \right) + 2\left( 0 \right) \cr
& = - 1 \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^1 {\ln x} dx = - 1 \cr} $$