Some conjectural formulas for $\pi$ are given in:
For example (I expressed an infinite sum given in the paper in terms of hypergeometric functions):
$$224 \,
_5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3
};1,1,1,1;8235 \sqrt{5}-18414\right)\\-100 \sqrt{5} \,
_5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3
};1,1,1,1;8235 \sqrt{5}-18414\right)\\-1655540 \,
_5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
};1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+740380
\sqrt{5} \,
_5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
};1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\-1237563
\,
_5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
};2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+553455
\sqrt{5} \,
_5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
};2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\stackrel{?}{=}\frac{4}{\pi^2}$$
You can also look at:
And one more from
J. Guillera homepage:
$$\sum_{n=0}^\infty\frac{(5418n^2+693n+29)(6n)!}{(-23887872000)^n n!^6}\stackrel?=\frac{128\sqrt5}{\pi^2}.$$
Following the same approach here:
$$I(a)=\int_0^1\frac{\ln x\ln^a(1-x)}{x^3}dx=\int_0^1\frac{\ln(1-x)\ln^ax}{(1-x)^3}dx\\\overset{IBP}{=}\frac12\int_0^1\frac{\ln^ax}{(1-x)^3}dx-\frac12a\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx$$.
The first integral:
$$\int_0^1\frac{\ln^ax}{(1-x)^3}dx=\frac12\sum_{n=2}^\infty n(n-1)\int_0^1 x^{n-2}\ln^ax\ dx\\=\frac12(-1)^aa!\sum_{n=2}^\infty\frac{n(n-1)}{(n-1)^{a+1}}=\frac12(-1)^aa!\sum_{n=1}^\infty\frac{n+1}{n^a}\\=\boxed{\frac12(-1)^aa!(\zeta(a-1)+\zeta(a))}$$
The second integral:
Use the partial fraction decomposition: $\frac1{(1-x)^2x}=\frac1x+\frac1{1-x}+\frac1{(1-x)^2}$
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx\\=\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx$$
where the first integral:
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1 x^{n-1}\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac1{n^{a+1}}=(-1)^a(a-1)!\zeta(a+1)$$
and the second integral
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac{H_n}{(n+1)^a}\\=(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}-\zeta(a+1)\right)$$
As for the third integral, its already calculated in the link I mentioned above which shows
$$\int_0^1\frac{\ln x\ln^b(1-x)}{x^2}dx=\int_0^1\frac{\ln(1-x)\ln^bx}{(1-x)^2}dx=(-1)^b b!\left(\zeta(b)-\sum_{n=1}^\infty\frac{H_n}{n^b}\right)$$
then we can write
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx=(-1)^a (a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)$$
Collect the three sub-integrals to get
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx=\boxed{(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}+\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)}$$
Combine the boxed results we get
$$I(a)=\frac14(-1)^aa!\left(3\zeta(a-1)+\zeta(a)-2\sum_{n=1}^\infty\frac{H_n}{n^a}-2\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}\right)$$
where $$\sum_{n=1}^\infty \frac{H_n}{n^a}=\left(1+\frac a2\right)\zeta(a+1)-\frac12\sum_{k=1}^{a-2}\zeta(k+1)\zeta(a-k)$$ is the generalized Euler identity.
Best Answer
Here is an approach without using the derivative of beta function :
$$I(b)=\int_0^1\frac{\ln x\ln^b(1-x)}{x^2}dx\overset{IBP}{=}\int_0^1\frac{\ln^b(1-x)}{x^2}dx-b\int_0^1\frac{\ln x\ln^{b-1}(1-x)}{x(1-x)}dx$$
where
$$\int_0^1\frac{\ln^b(1-x)}{x^2}dx=\int_0^1\frac{\ln^bx}{(1-x)^2}dx=\sum_{n=1}^\infty n\int_0^1 x^{n-1}\ln^bx\ dx\\=(-1)^b b!\sum_{n=1}^\infty\frac{n}{n^{b+1}}=\boxed{(-1)^b b!\zeta(b)}$$
and
$$\int_0^1\frac{\ln x\ln^{b-1}(1-x)}{x(1-x)}dx=\int_0^1\frac{\ln(1-x)\ln^{b-1}x}{x(1-x)}dx\\=-\sum_{n=1}^\infty H_n\int_0^1 x^{n-1}\ln^{b-1}x \ dx=\boxed{-(-1)^{b-1}(b-1)!\sum_{n=1}^\infty\frac{H_n}{n^b}}$$
Combine the two results we have
$$I(b)=(-1)^b b!\left(\zeta(b)-\sum_{n=1}^\infty\frac{H_n}{n^b}\right)$$
where the generalized Euler sum:
$$\sum_{n=1}^\infty \frac{H_n}{n^b}=\left(1+\frac b2\right)\zeta(b+1)-\frac12\sum_{k=1}^{b-2}\zeta(k+1)\zeta(b-k)$$