Answer
$$ - \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x\ln x} dx \cr
& x\ln x{\text{ is not defined at }}x = 0,{\text{so we can write}} \cr
& \int_0^1 {x\ln x} dx = \mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {x\ln x} dx \cr
& {\text{Integrate }}\int {x\ln x} dx{\text{ by parts}} \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx,{\text{ }}dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& \int {udv} = uv - \int {vdu} = \frac{{{x^2}}}{2}\ln x - \int {\frac{{{x^2}}}{2}\left( {\frac{1}{x}} \right)dx} \cr
& = \frac{{{x^2}}}{2}\ln x - \frac{1}{4}{x^2} + C \cr
& \cr
& \mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {x\ln x} dx = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {\frac{{{x^2}}}{2}\ln x - \frac{1}{4}{x^2}} \right]_b^1 \cr
& = \mathop {\lim }\limits_{b \to {0^ + }} \left[ {\frac{{{1^2}}}{2}\ln 1 - \frac{1}{4}{1^2}} \right] - \mathop {\lim }\limits_{b \to {0^ + }} \left[ {\frac{{{b^2}}}{2}\ln b - \frac{1}{4}{b^2}} \right] \cr
& = \mathop {\lim }\limits_{b \to {0^ + }} \left( { - \frac{1}{4}} \right) - \mathop {\lim }\limits_{b \to {0^ + }} \left[ {\frac{{{b^2}}}{2}\ln b - \frac{1}{4}{b^2}} \right] \cr
& {\text{Evaluate the limit when }}b \to {0^ + } \cr
& = - \frac{1}{4} - \left[ {0 - \frac{1}{4}\left( 0 \right)} \right] \cr
& = - \frac{1}{4} \cr
& {\text{The following graph confirms the result}} \cr} $$
