Xrfgtjtk

We start by introducing the integral
$$
I(a,b)=frac{1}{2}int_0^1y^{a-1}(1-y)^b,dy=frac{1}{2}B(a,1+b),
$$
where $B$ denotes the beta function. Note that this integral is singular at $a=0$ and $b=-1$. Since $partial_a y^a=y^aln y$ we are
led to calculate
$$
partial_{a,b,b}I(a,b)=frac{1}{2}int_0^1 y^{a-1}(1-y)^bln ybigl(ln(1-y)bigr)^2,dy
$$
as $a$ and $b$ tend to $0$. We will below inser the "non-dangerous" point $b=0$. In other words, we want to calculate
$$
partial_{a,b,b}B(a,1+b)mid_{ato 0^+,bto 0}.
$$
When differentiating the beta function, polygammas appear. Indeed,
$$
begin{aligned}
partial_bB(a,1+b)&=B(a,1+b)bigl(psi_0(1+b)-psi_0(1+a+b)bigr)\
partial_{b,b}B(a,1+b)&=B(a,1+b)Bigl(bigl(psi_0(1+b)-psi_0(1+a+b)bigr)^2
+psi_1(1+b)-psi_1(1+a+b)Bigr).
end{aligned}
$$
Next, we can actually insert $b=0$ before we differentiate with respect to
$a$ and take the limit $ato 0$. We should differentiate the function (here we have used the facts that $psi_0(1)=-gamma$ (Euler's constant) and that $psi_1(1)=pi^2/6$)
$$
f(a)=B(a,1)Bigl(bigl(gamma+psi_0(1+a)bigr)^2+frac{pi^2}{6}-psi_1(1+a)Bigr)
$$
and calculate $lim_{ato 0^+}f'(a)$. We get that
$$
begin{aligned}
f'(a)&=B(a,1)bigl(psi_0(a)-psi_0(1+a)bigr)Bigl(bigl(gamma+psi_0(1+a)bigr)^2+frac{pi^2}{6}-psi_1(1+a)Bigr)\
&quad+B(a,1)Bigl(2bigl(gamma+psi_0(1+a)bigr)psi_1(1+a)-psi_2(1+a)Bigr)
end{aligned}
$$
Next, we use the (non-obvious) expansions around $a=0$
$$
begin{aligned}
B(a,1)&=frac{1}{a}+O(1)\
psi_0(a)&=-frac{1}{a}-gamma+O(a)\
psi_0(1+a)&=-gamma+frac{pi^2}{6}a+O(a^2)\
psi_1(1+a)&=frac{pi^2}{6}+psi_2(1)a+frac{pi^4}{30}a^2+O(a^3)\
psi_2(1+a)&=psi_2(1)+frac{pi^4}{15}a+O(a^2).
end{aligned}
$$
to find that, as $ato0^+$,
$$
begin{aligned}
f'(a)&approx -frac{1}{a^2}Bigl(bigl(frac{pi^2}{6}abigr)^2-psi_2(1)a-frac{pi^4}{30}a^2Bigr)+frac{1}{a}Bigl(2frac{pi^2}{6}afrac{pi^2}{6}-psi_2(1)-frac{pi^4}{15}aBigr)+O(a)\
&=-frac{pi^4}{180}+O(a)
end{aligned}
$$
as $ato 0^+$. We conclude that
$$
partial_{a,b,b}B(a,1+b)mid_{ato 0^+,bto 0}=-frac{pi^4}{180}.
$$
Finally, dividing by $2$ (remember, we had a one-half in front of the beta function in the beginning), we get that
$$
int_0^1frac{1}{2y}ln y(ln(1-y))^2,dy=-frac{1}{360}pi^4.
$$

There is a closed form antiderivative that can be found using repeated integration by parts:
$$begin{align}intfrac{ln ycdotln^2(1-y)}{2y}dy=frac{ln^4(1-y)}8+frac{ln ycdotln ^3(1-y)}6+left(frac{pi^2}{12}-frac{ln ^2y}2right)cdotln ^2(1-y)\
+left[left(frac{pi^2}6+operatorname{Li}_2left(tfrac y{y-1}right)right)cdotln y-operatorname{Li}_3(1-y)-operatorname{Li}_3left(tfrac y{y-1}right)right]cdotln(1-y)\
+left[vphantom{Large|}zeta(3)-operatorname{Li}_3(1-y)right]cdotln y+operatorname{Li}_4(1-y)-operatorname{Li}_4(y)-operatorname{Li}_4left(tfrac y{y-1}right)color{gray}{+C}end{align}$$

Another one:

Write $log^2(1-y)=sum_{n,m=1}^{infty}frac{y^{m+n}}{mn}$

we obtain (exchanging summation and integration)

$$
2I=sum_{n,m=1}^{infty}frac{1}{mn}int_0^1log(y)y^{m+n-1}=\
-underbrace{sum_{n,m=1}^{infty}frac{1}{mn}frac{1}{(m+n)^2}}_{S}
$$

the double sum can be tackeled by writing (i shamelessly benefit from this awesome answer)

$$
S=sum_{n,m=1}^{infty}frac{1}{mn}frac{1}{(m+n)^2}=frac{1}{2}sum_{n,m=1}^{infty}frac{1}{m^2n^2}-frac{1}{2}sum_{n,m=1}^{infty}frac{1}{(m+n)^2n^2}-frac{1}{2}sum_{n,m=1}^{infty}frac{1}{(m+n)^2m^2}
$$

shifting arguments in the last two sums gives

$$
2S=sum_{n,m=1}^{infty}frac{1}{mn}frac{1}{(m+n)^2}-sum_{m=1,n<m}^{infty}frac{1}{n^2m^2}-sum_{m=1,n>m}^{infty}frac{1}{n^2m^2}
$$

which yields

$$
2S=sum_{n=m=1}^{infty}frac{1}{m^2n^2}=zeta(4)=frac{pi^4}{90}
$$

and therefore

$$
I=-S/2=-frac{pi^4}{360}
$$

This question, for some reason, popped up in the Top Questions tab, and I thought I'd share another way to solve this integral using Harmonic Numbers. I hope you guys don't mind!

First, we use integration by parts on $u=log^2(1-x)$ to get$$begin{align*}intlimits_0^1dx,frac {log xlog^2(1-x)}{x} & =frac 12log^2xlog^2(1-x),Biggrrvert_0^1+intlimits_0^1dx,frac {log^2xlog(1-x)}{1-x}\ & =intlimits_0^1dx,frac {log^2xlog(1-x)}{1-x}end{align*}$$Now use the fact that

$$H(x)=-frac {log(1-x)}{1-x}=sumlimits_{ngeq1}H_nx^n$$

And substitute to get$$begin{align*}intlimits_0^1dx,frac {log xlog^2(1-x)}{x} & =-sumlimits_{ngeq1}H_nintlimits_0^1dx,x^nlog^2x\ & =-limlimits_{muto0}sumlimits_{ngeq1}H_nfrac {partial^2}{partialmu^2}frac 1{n+mu+1}\ & =-2sumlimits_{ngeq1}frac {H_n}{(n+1)^3}end{align*}$$Split the sum up and use a well-known formula due to Euler

$$sumlimits_{ngeq1}frac {H_n}{n^m}=frac 12(m+2)zeta(m+1)-frac 12sumlimits_{n=1}^{m-2}zeta(m-n)zeta(n+1)$$

Therefore$$I=2zeta(4)-5zeta(4)+zeta^2(2)=-frac {pi^4}{180}$$Our desired integral is half of that, so take half and the answer is$$intlimits_0^1dx,frac {log xlog^2(1-x)}{2x}color{blue}{=-frac {pi^4}{360}}$$

I offer up yet another approach, this one relying on the Maclaurin series expansion for $ln^2 (1 - x)$. It is similar to that used by @Frank W.

As was shown here
$$ln^2 (1 - x) = 2 sum_{n = 2}^infty frac{H_{n - 1} x^n}{n}, qquad |x| < 1.$$
Here $H_n$ denotes the Harmonic number. So for the integral we may write
begin{align}
int_0^1 frac{1}{2x} ln x ln^2 (1 - x) , dx &= sum_{n = 2}^infty frac{H_{n - 1}}{n} int_0^1 x^{n - 1} ln x , dx = -sum_{n = 2}^infty frac{H_{n - 1}}{n^3},
end{align}
after integrating by parts. From properties for the Harmonic number, since
$$H_n = H_{n - 1} + frac{1}{n},$$
the infinite sum can be rewritten as
begin{align}
int_0^1 frac{1}{2x} ln x ln^2 (1 - x) , dx &= sum_{n = 2}^infty frac{1}{n^4} - sum_{n = 2}^infty frac{H_n}{n^4} = sum_{n = 1}^infty frac{1}{n^4} - sum_{n = 1}^infty frac{H_n}{n^4}.
end{align}
Values for each of these sums are well known. For the first
$$sum_{n = 1}^infty frac{1}{n^4} = zeta (4) = frac{pi^4}{90}.$$
For the second sum
$$sum_{n = 1}^infty frac{H_n}{n^4} = frac{pi^4}{72}.$$
(for various proofs of this result, see here) Thus
$$int_0^1 frac{1}{2x} ln x ln^2 (1 - x) , dx = frac{pi^4}{90} - frac{pi^4}{72} = -frac{pi^4}{360}.$$

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newcommand{ds}[1]{displaystyle{#1}}
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newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
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$ds{{1 over 2}int_{0}^{1}{lnpars{y}ln^{2}pars{1 - y} over y},dd y =
-,{pi^{4} over 360}:
{large ?}}$.

begin{align}
&bbox[10px,#ffd]{ds{{1 over 2}int_{0}^{1}{lnpars{y}ln^{2}pars{1 - y} over y},dd y}} =
left.{1 over 2},{partial^{3} over partialnu^{2}partialmu}int_{0}^{1}{y^{mu}bracks{pars{1 - y}^{nu} - 1} over y},dd y,rightvert_{ {largemu = 0^{+}} atop {largenu = 0}}
\[5mm] = &
{1 over 2},{partial^{3} over partialnu^{2}partialmu}
bracks{{Gammapars{mu}Gammapars{nu + 1} over Gammapars{mu + nu + 1}} - {1 over mu}}_{ {largemu = 0^{+}} atop {largenu = 0}}quad
pars{~Gamma: Gamma Function~}
\[5mm] = &
{1 over 2},{partial^{3} over partialnu^{2}partialmu}
bracks{{pi over Gammapars{1 - mu}sinpars{pimu}},{Gammapars{nu + 1} over Gammapars{mu + nu + 1}} - {1 over mu}}_{ {largemu = 0^{+}} atop {largenu = 0}}
\[5mm] = &
{1 over 2},{partial^{3} over partialnu^{2}partialmu}
bracks{{1 over mu},{Gammapars{nu + 1} over Gammapars{1 - mu}Gammapars{mu + nu + 1}} + {pi^{2} over 6},mu}
_{ {largemu = 0^{+}} atop {largenu = 0}}
\[5mm] = &
{1 over 2},{partial^{3} over partialnu^{2}partialmu}
bracks{left.{1 over 2},partiald[2]{}{x}{Gammapars{nu + 1} over Gammapars{1 - x}Gammapars{x + nu + 1}},rightvert_{ x = 0^{+}}mu + {pi^{2} over 6},mu}_{ {largemu = 0^{+}} atop {largenu = 0}}
\[5mm] = &
{1 over 2},{partial^{2} over partialnu^{2}}
bracks{left.{1 over 2},partiald[2]{}{x}{Gammapars{nu + 1} over Gammapars{1 - x}
Gammapars{x + nu + 1}},rightvert_{ x = 0^{+}} + {pi^{2} over 6}}
_{ {largemu = 0^{+}} atop {largenu = 0}} =
{1 over 4},{partial^{4} over partialnu^{2}partialmu^{2}}
{nu choose mu + nu}_{ {largemu = 0^{+}} atop {largenu = 0}}
\[5mm] = &
{1 over 4},partiald[2]{}{nu}
bracks{-,{pi^{2} over 6} + H^{2}_{nu} - Psi, 'pars{1 + nu}}
_{ nu = 0}quad
pars{~H_{z}: Harmonic Number~}
\[5mm] = &
{1 over 4}bracks{2Psi, '^{2}pars{1} + 2H_{0},Psi,''pars{1}- Psi, '''pars{1}}
qquadqquadqquadqquad
left{begin{array}{lcr}
ds{Psi, 'pars{1}} & ds{=} & ds{pi^{2} over 6}
\
ds{Psi, '''pars{1}} & ds{=} & ds{pi^{4} over 15}
\
ds{H_{0}} & ds{=} & ds{0}
end{array}right.
\[5mm] = &
{1 over 4}bracks{2pars{pi^{2} over 6}^{2} + 0 - {pi^{4} over 15}} =
bbx{-,{pi^{4} over 360}}
end{align}