for integration experts

The result you have got is the best you can do (except that the factor in the beginning will be 2 and not 1/2). The sum
\sum_{k=1}^\infty \dfrac{1}{k^3}
is called Apery's constant which is basically \zeta(3)\approx 1.202 where \zeta(x) is the Riemann zeta function.

oh ok actually i started from the Apery's constant and found this integration ,i thought if its solved by someone here , there would be a breakthrough !

also sir can u tell me how where u able to derive aperys constant from this integration ?

i tried hundreds of methods but was unable to solve :(

Let
I=\int_0^1 \dfrac{\ln x \ln(1-x)}{x(1-x)}\ \mathrm{d}x
Rewrite it as
I=\int_0^1 (x+1-x)\dfrac{\ln x \ln(1-x)}{x(1-x)}\ \mathrm{d}x
=\int_0^1 \dfrac{\ln x \ln(1-x)}{1-x}\ \mathrm{d}x+ \int_0^1\dfrac{\ln x \ln(1-x)}{x}\right)\ \mathrm{d}x
=2\int_0^1\dfrac{\ln x \ln(1-x)}{x}\right)\ \mathrm{d}x
where I used the property of definite integral that \int_a^b f(x) \mathrm{d}x=\int_a^b f(a+b-x)\ \mathrm{d}x
Now use the series expansion of \ln (1-x)=-\sum_{k=1}^\infty \dfrac{x^k}{k}
to obtain that
I=-2\int_0^1 \ln x\left(\sum_{k=1}^\infty \dfrac{x^{k-1}}{k}\right)\ \mathrm{d}x = -2\sum_{k=1}^\infty\dfrac{1}{k}\int_0^1x^{k-1}\ln x\ \mathrm{d}x
Finally, the integral
\int_0^1x^{k-1}\ln x\ \mathrm{d}x = \left|\dfrac{x^k\ln x}{k}\right|_0^1-\dfrac{1}{k}\int_0^1x^{k-1}\ \mathrm{d}x
The first term evaluates to zero and so we are left with
\int_0^1x^{k-1}\ln x\ \mathrm{d}x = -\dfrac{1}{k^2}
Hence, we obtain
\boxed{I=2\sum_{k=1}^\infty\dfrac{1}{k^3}}