How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$ $$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$
This number looks like $\zeta(5)$ value. 
We expand the terms
$$\int_0^1\frac{\frac{\pi^2}{6}}{1-x}\cdot \ln^2(x) \, \mathrm dx-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x} \cdot \ln^2(x) \, \mathrm dx$$
$$2\zeta(2)\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$$
this last integral it is very complicate to compute...
Can any user please help to show that whether this value is equal to $\zeta(5)$ or not?  
 A: Start with $1-x\mapsto x$
$$I=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2x}{1-x}dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2(1-x)}{x}dx$$
By Cauchy product we have 
$$\ln(1-x)\operatorname{Li}_2(x)=-\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$
multiply both sides by $\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $x=1$ and use the fact that $-\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$ we get
$$I=\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)\left(\frac{H_n}{n}\right)$$
$$=2\sum_{n=1}^\infty\frac{H_n^2}{n^3}+\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}-3\sum_{n=1}^\infty\frac{H_n}{n^4}$$
For the first sum, its evaluated here
$$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$$ 
The second sum, can be found here
$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$
and the well-known result 
$$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
Combine the three sums to get
$$\boxed{I=2\zeta(2)\zeta(3)-\zeta(5)}$$

A different approach is to apply integration by parts to your original integral before breaking the inegrand.
A: We can have a nice generalization for
$$I_n=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(1-x)}{1-x}\ln^2x\ dx=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(x)}{x}\ln^2(1-x)\ dx$$

From $$\ln^2(1-x)=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}x^k$$
It follows that
$$I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{k-1}(\zeta(n)-\operatorname{Li}_n(x))\ dx$$
By integration by parts we have
$$\int_0^1x^{k-1}\operatorname{Li}_n(x)\ dx=(-1)^{n-1}\frac{H_k}{k^n}-\sum_{i=1}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}$$
$$\Longrightarrow I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(n)}{k}+(-1)^{n}\frac{H_k}{k^n}+\sum_{i=\color{red}{1}}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}\right)$$
or
$$I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{i=\color{red}{2}}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}+(-1)^{n}\frac{H_k}{k^n}\right),\quad n=2,3,...$$

Applications
$$I_2=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{H_k}{k^2}\right)$$
$$ I_3=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(2)}{k^2}-\frac{H_k}{k^3}\right)$$
$$I_4=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(3)}{k^2}-\frac{\zeta(2)}{k^3}+\frac{H_k}{k^4}\right)$$
$$I_5=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(4)}{k^2}-\frac{\zeta(3)}{k^3}+\frac{\zeta(2)}{k^4}-\frac{H_k}{k^5}\right)$$
A: Here is an easier way, let $1-x\mapsto x$
$$I=\int_0^1\frac{\zeta(2)-\operatorname{Li}_2(1-x)}{1-x}\ln^2x\ dx=\int_0^1\frac{\zeta(2)-\operatorname{Li}_2(x)}{x}\ln^2(1-x)\ dx$$
Now use $\ln^2(1-x)=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$ we get
$$I=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^1 x^{n-1}(\zeta(2)-\operatorname{Li}_2(x))\ dx$$
$$=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(\frac{\zeta(2)}{n}-\frac{\zeta(2)}{n}+\frac{H_n}{n^2}\right)$$
$$=2\sum_{n=1}^\infty\frac{H_n^2}{n^3}-2\sum_{n=1}^\infty\frac{H_n}{n^4}$$
These two sums are mentioned in my previous solution above, collecting them gives $ I=\zeta(5)$
A: Here is an alternative approach through the harmonic sum thicket.
Starting with Euler's reflexion formula for the dilogarithm function, namely
$$\operatorname{Li}_2 (x) + \operatorname{Li}_2 (1 - x) = \zeta (2) - \ln x \ln (1- x),$$
your integral $I$ can be rewritten as
$$I = \int_0^1 \frac{\ln^3 x \ln (1 - x)}{1 - x} \, dx + \int_0^1 \frac{\ln^2 x \operatorname{Li}_2 (x)}{1 - x} \, dx = I_1 + I_2.$$
For the first integral, from the generating function for the harmonic numbers $H_n$ we have
$$\frac{\ln (1 - x)}{1 - x} = - \sum_{n = 1}^\infty H_n x^n.$$
Thus
\begin{align}
I_1 &= -\sum_{n = 1}^\infty H_n \int_0^1 x^n \ln^3 x \, dx\\
&= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\
&= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= 6 \sum_{n = 1}^\infty \frac{H_n}{(n + 1)^4} = 6 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^4},
\end{align}
after reindexing $n \mapsto n - 1$. Since
$$H_n = H_{n - 1} + \frac{1}{n},$$
this leads to
$$I_1 = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \sum_{n = 1}^\infty \frac{1}{n^2} = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \zeta (5).$$
For the second integral, from the generating function for the generalised harmonic numbers of order two $H^{(2)}_n$ we have
$$\frac{\operatorname{Li}_2 (x)}{1 - x} = \sum_{n = 1}^\infty H^{(2)}_n x^n.$$
Thus
\begin{align}
I_2 &= \sum_{n = 1}^\infty H^{(2)}_n \int_0^1 x^n \ln^2 x \, dx\\
&= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\
&= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= 2\sum_{n = 1}^\infty \frac{H^{(2)}_n}{(n + 1)^3} = 2 \sum_{n = 2}^\infty \frac{H^{(2)}_{n-1}}{n^3},
\end{align}
after reindexing $n \mapsto n - 1$. Since
$$H^{(2)}_n = H^{(2)}_{n - 1} + \frac{1}{n^2},$$
we have
$$I_2 = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \sum_{n = 1}^\infty \frac{1}{n^5} = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \zeta (5).$$
Returning to our integral, we have
$$I = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 8 \zeta (5).$$
Since
$$\sum_{n = 1}^\infty \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3),$$
and
$$\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} = 3 \zeta (2) \zeta (3) - \frac{9}{2} \zeta (5),$$
it is immedate that $I = \zeta (5)$, as desired. 
