What is $\int_{x}^{1}-y\log(1-y)\,dy$? 
On a stock $1$ meter long is casually marked a point $X \sim U[0,1]$. Let $X=x$, is also marked a second point $Y\sim U[x,1]$.


1) Find the density of $(X,Y)$ showing the domain.

$\rightarrow \quad f_{XY}(x,y)=\frac{1}{1-x}\mathbb{I}_{[0,1]}(x)\mathbb{I}_{[x<y<1]}(y)$

2) Say if $X$ and $Y$ are independent or not, and compute $\operatorname{Cov}(X,Y)$.

$\rightarrow \quad  \begin{array}[t]{l}
f_Y(y)=-\log(1-y)\mathbb{I}_{[0,1]}(y) \\
\Rightarrow \quad f_X(x)f_Y(y)\neq f_{XY}(x,y) \\
\Rightarrow \quad \text{$X$ and $Y$ are not independent}
\end{array}$
$\rightarrow \quad $ $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$, where
$$
\mathbb{E}[Y]=\int_{x}^{1}-y\log(1-y) \, dy.
$$
Fixing $t=1-y$,
$$
\Rightarrow \quad  \int_{1-x}^{0}-(1-t)\log(t) \, (-dt) = \int_{0}^{1-x}(t-1)\log(t) \, dt.
$$
Then, for $f(x)=\log(t)$, $g'(x)=(t-1)$, $g(x)=\frac{1}{2}(t-1)^2$, I have
$$
\left.\frac{1}{2}(t-1)^2\log(t)\right|_{0}^{1-x} - \frac{1}{2}\int_{0}^{1-x}\frac{(t-1)^2}{t} \, dt,
$$
but obviously $\log(0)=-\infty$. How can I solve that integral?

3) Now we assume to break the stick in the points $X$ and $Y$, and to form a triangle with the pieces that we have. Remembering that in a triangle the sum of the lengths of two sides must be greater than the length of the third side, what is the probability to form a triangle with the three pieces of the stick?

For this last point, I still have to think about it. Later I'll edit.

Thanks in advance for any help.
EDIT: First of all, thank you all for your answers.
Said that, seen the result (complicated, actually), I thought to solve the second point as follows:
$Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$ where

*

*$\mathbb{E}[X]:=\frac{1}{2}$;


*$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\frac{1}{2}\mathbb{E}[x+1]=\frac{3}{4}$;


*$\mathbb{E}[XY]=\frac{1}{2}\int_{0}^{1}x\mathbb{E}[Y|X]dx=\frac{5}{24}$;
Thus $Cov(X,Y)=\frac{5}{24}-\frac{1}{2}\cdot \frac{3}{4}=-\frac{1}{6}$.
 A: With the choice $$u = -\log(1-y), \quad du = \frac{1}{1-y} \, dy, \\ dv = y \, dy, \quad v = \frac{y^2}{2},$$ we obtain $$\int -y \log(1-y) \, dy = -\frac{y^2}{2} \log (1-y) - \frac{1}{2} \int \frac{y^2}{1-y} \, dy.$$  Then a substitution of the form $$t = 1-y, \quad dt = -dy$$ gives $$\int \frac{y^2}{1-y} \, dy = - \int \frac{(1-t)^2}{t} \, dt = - \int \frac{1}{t} - 2 + t \, dt = -\log t + 2t - \frac{t^2}{2} + C.$$  Consequently $$\int -y \log(1-y) \, dy = \frac{1-y^2}{2} \log (1-y) - (1-y) + \frac{(1-y)^2}{4} + C.$$  Evaluating this at $y = 1$, noting that $$\lim_{z \to 0} z \log z = 0,$$ we get  $$\int_{y=x}^1 -y \log (1-y) \, dy = (1-x) - \frac{(1-x)^2}{4} - \frac{1-x^2}{2} \log (1-x).$$  This may be factored as $$\frac{1-x}{4}\bigl(3+x - 2(1+x)\log(1-x)\bigr).$$
A: $\begin{array}\\
I(x)
&=\int_{x}^{1}-y\log(1-y)\,dy\\
&=\int_{1-x}^{0}-(1-y)\log(y)\,d(1-y)
\qquad y \to 1-y\\
&=\int^{1-x}_{0}-(1-y)\log(y)\,dy\\
&=-\int^{1-x}_{0}\log(y)\,dy+\int^{1-x}_{0}y\log(y)\,dy\\
&=-(y\log(y)-y)|^{1-x}_{0}+(\frac12 y^2\log(y)-\frac14 y^2|^{1-x}_{0}\\
&=-((1-x)\log(1-x)-1+x)+(\frac12 (1-x)^2\log(1-x)-\frac14 (1-x)^2\\
&=-(1-x)\log(1-x)+1-x+\frac12 (1-x)^2\log(1-x)-\frac14 (1-x)^2\\
&=(-(1-x)+\frac12(1-x)^2)\log(1-x)+(1-x)(1-\frac14(1-x))\\
&=(1-x)(-1+\frac12(1-x))\log(1-x)+(1-x)(\frac34+\frac14 x)\\
&=(1-x)(-\frac12(1+x))\log(1-x)+(1-x)(\frac34+\frac14 x)\\
&=(1-x)\left((-\frac12(1+x))\log(1-x)+(\frac34+\frac14 x)\right)\\
\end{array}
$
A: $$\int_x^1y\ln(1-y)dy=\sum_{n=1}^\infty\frac1n\int_x^1y^{n+1}dy=\sum_{n=1}^\infty\frac1n\left(\frac{1}{n+2}-\frac{x^{n+2}}{n+2}\right)$$
$$=\sum_{n=1}^\infty\frac{1}{n(n+2)}-\sum_{n=1}^\infty\frac{x^{n+2}}{n(n+2)}$$
$$=\frac12H_2-\frac12\sum_{n=1}^\infty\frac{x^{n+2}}{n}+\frac12\sum_{n=1}^\infty\frac{x^{n+2}}{n+2}$$
$$=\frac34-\frac12\left(-x^2\ln(1-x)\right)+\frac12\left(-x-\frac12x^2-\ln(1-x)\right)$$
$$=\frac12x^2\ln(1-x)-\frac12\ln(1-x)-\frac12x^2-\frac12x+\frac34$$
A: \begin{align}
EY&=\int_0^1\int_x^1\frac y{1-x}dydx\\
&=\frac12\int_0^1\frac{1-x^2}{1-x}dx=\frac34\\
EXY&=\int_0^1\int_x^1\frac {xy}{1-x}dydx\\
&=\frac12\int_0^1\frac {x(1-x^2)}{1-x}dy=\frac5{12}\\
Cov(X,Y)&=EXY-EXEY\\
&=\frac5{12}-\frac12 \times \frac34
\end{align}
