I am reqired to evaluate 
$$\int\_0^1 x\log x dx$$
I am getting 
$$\frac{x^2}2\log x-\frac{x^2}4$$
from $0$ to $1$ which means $$\frac12\log1-\frac14-0\log0-0 = \frac{-1}{4}-0\log0.$$
The answer in the book is $\frac{-1}4$. Can the second part containing $\log0$, which is indeterminate, be neglected?