How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$

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

$$\boxed{I=2\zeta(2)\zeta(3)-\zeta(5)}$$

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A different approach is to apply integration by parts to your original integral before breaking the inegrand.

We can have a nice generalization for

$$I_n=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(1-x)}{1-x}\ln^2x\ dx=\int_0^1\frac{\zeta(n)-\operatorname{Li}_n(x)}{x}\ln^2(1-x)\ dx$$

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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

$$I_n=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{i=\color{red}{2}}^{n-1}(-1)^i\frac{\zeta(n-i+1)}{k^i}+(-1)^{n}\frac{H_k}{k^n}\right),\quad n=2,3,...$$

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Applications

$$I_2=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{H_k}{k^2}\right)$$

$$ I_3=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(2)}{k^2}-\frac{H_k}{k^3}\right)$$

$$I_4=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(3)}{k^2}-\frac{\zeta(2)}{k^3}+\frac{H_k}{k^4}\right)$$

$$I_5=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\frac{\zeta(4)}{k^2}-\frac{\zeta(3)}{k^3}+\frac{\zeta(2)}{k^4}-\frac{H_k}{k^5}\right)$$

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)$