Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^e {\ln {x^2}} dx \cr
& {\text{Use logarithmic properties}} \cr
& = \int_0^e {2\ln x} dx \cr
& = 2\int_0^e {\ln x} dx \cr
& \ln x{\text{ is not defined at }}x = 0,{\text{so we can write}} \cr
& 2\int_0^e {\ln x} dx = 2\mathop {\lim }\limits_{b \to {0^ + }} \int_b^e {\ln x} dx \cr
& {\text{Integrate }}\int {\ln x} dx{\text{ by parts}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx,{\text{ }}dv = dx,{\text{ }}v = x \cr
& \int {udv} = uv - \int {vdu} = x\ln x - \int {x\left( {\frac{1}{x}} \right)dx} \cr
& = x\ln x - x + C \cr
& \cr
& 2\mathop {\lim }\limits_{b \to {0^ + }} \int_b^e {\ln x} dx = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {x\ln x - x} \right]_b^e \cr
& = 2\mathop {\lim }\limits_{b \to {0^ + }} \left[ {e\ln e - e} \right] - 2\mathop {\lim }\limits_{b \to {0^ + }} \left[ {b\ln b - b} \right] \cr
& = 2\mathop {\lim }\limits_{b \to {0^ + }} \left( 0 \right) - 2\mathop {\lim }\limits_{b \to {0^ + }} \left[ {b\ln b - b} \right] \cr
& {\text{Evaluate the limit when }}b \to {0^ + } \cr
& = - 2\left( 0 \right) - 2\left[ {0 - \left( 0 \right)} \right] \cr
& = 0 \cr
& {\text{The following graph confirms the result}} \cr} $$
