Double sum that turns out be harmonic

Here I wish to share my solution

Note that $\sum_{n=3}^{\infty}\dfrac{2}{(n+1)(n+2)} =\dfrac{2}{4}=\dfrac{1}{2} \\ \sum_{n=3}^{\infty}\dfrac{3}{n(n+1)(n+2)} =\sum_{n=1}^{\infty}\dfrac{3}{n(n+1)(n+2)}\\-\dfrac{3}{6}-\dfrac{3}{24}=\dfrac{3}{2\cdot 2!} -\dfrac{5}{8}=\dfrac{1}{8}$

Note that $\dfrac{1}{nm^2+n^2m+2mn} =\dfrac{1}{mn(m+n+2)}=\dfrac{1}{mn}\int_0^1 x^{m+n+1}\,dx$ then we have $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{mn} \int_0^1 x^{m+n+1}\,dx=\int_0^1x\left(\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{x^{m}\cdot x^n}{mn}\right)\,dx \\= \int_0^1 x\ln^2(1-x)\,dx =\int_0^1 (1-x)\ln^2 x\,dx$ hence integrating we have the result $\dfrac{7}{4}$