$\int_0^1 x(1-x)\log(x(1-x))dx=?$ I would like to compute the following integral. The answer must be $-\frac{5}{18}$ but I do not know how to evaluate. 
$$\int_0^1 x(1-x)\log(x(1-x))dx$$
Thank you for your help in advance. 
 A: Hint:
$$\int_0^1 x^k \log{x} = -\frac{1}{(k+1)^2}$$
Taylor expand the $\log(1-x)$ piece about $x=0$.
$$\begin{align}\int_0^1 x(1-x)\log(x(1-x))dx &= \int_0^1 x(1-x)\log{x} + \int_0^1 x(1-x)\log{(1-x)}\\&= -\frac{1}{4} + \frac{1}{9} - \sum_{k=1}^{\infty} \frac{1}{k}\int_0^1 dx \: x(1-x) x^k \\ &= -\frac{5}{36} - \sum_{k=1}^{\infty} \frac{1}{k} \left (\frac{1}{k+2}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+2} \right ) + \frac{1}{3} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \left ( 1 + \frac{1}{2} \right )+ \frac{1}{3} \left ( 1 + \frac{1}{2}+ \frac{1}{3} \right )\\ &= -\frac{5}{36} - \frac{3}{4}+ \frac{11}{18}\\\therefore \int_0^1 x(1-x)\log(x(1-x))dx  &= -\frac{5}{18} \end{align}$$
A: Use $\log(x(1-x)) = \log(x) + \log(1-x)$:
$$
   \int_0^1 x(1-x) \log(x(1-x)) \mathrm{d}x =  \int_0^1 x(1-x) \log(x) \mathrm{d}x + \int_0^1 x(1-x) \log(1-x) \mathrm{d}x
$$
Making change of variables $x \to 1-x$ in the second integral:
$$
  \int_0^1 x(1-x) \log(x(1-x)) \mathrm{d}x = 2 \int_0^1 x(1-x) \log(x) \mathrm{d}x
$$
Now, integrate by parts:
$$
  \int_0^1 x(1-x) \log(x) \mathrm{d}x = \left.\left(\frac{x^2}{2} - \frac{x^3}{3}\right) \log(x) + \frac{x^3}{9} - \frac{x^2}{4} \right|_{x \downarrow 0}^{x=1} = \frac{1}{9} - \frac{1}{4} = -\frac{5}{36}
$$
Combining we get $-5/18$ as claimed.
