<s>[INST] $\int_0^1 x(1-x)\log(x(1-x))dx=?$ 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. 
 [/INST] Hint:
$$\int_0^1 x^k \log{x} = -\frac{1}{(k+1)^2}$$
Taylor expand the $\log(1-x)$ piece about $x=0$.
$$\begin{align}\int_0^1 x(1-x)\log(x(1-x))dx &= \int_0^1 x(1-x)\log{x} + \int_0^1 x(1-x)\log{(1-x)}\\&= -\frac{1}{4} + \frac{1}{9} - \sum_{k=1}^{\infty} \frac{1}{k}\int_0^1 dx \: x(1-x) x^k \\ &= -\frac{5}{36} - \sum_{k=1}^{\infty} \frac{1}{k} \left (\frac{1}{k+2}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+2} \right ) + \frac{1}{3} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \left ( 1 + \frac{1}{2} \right )+ \frac{1}{3} \left ( 1 + \frac{1}{2}+ \frac{1}{3} \right )\\ &= -\frac{5}{36} - \frac{3}{4}+ \frac{11}{18}\\\therefore \int_0^1 x(1-x)\log(x(1-x))dx  &= -\frac{5}{18} \end{align}$$
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