Resembling Resemblence-4

Using integral representation $\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n} H_n = -\displaystyle\int_0^1 \sum_{n=1}^\infty (-x)^n H_n \dfrac{\mathrm{d} x }{x}$

Now:

$-\displaystyle\sum_{n=1}^\infty (-x)^n H_n = -\displaystyle\sum_{n=1}^\infty x^n \displaystyle\sum_{k=0}^{n-1} (-1)^k \frac{(-1)^{n-k}}{n-k}$ $= -\displaystyle\sum_{n=0}^\infty (-x)^n \cdot \displaystyle\sum_{k=1}^\infty \frac{(-x)^k}{k} = \frac{\log(1+x)}{1+x}$

Thus

$I=\color{#3D99F6}{\displaystyle\int_0^1 \dfrac{\log(1+x)}{1+x} \dfrac{\mathrm{d}x}{x} = \left. \left(-\dfrac{1}{2} \log^2(1+x) - \text{Li}_2(-x) \right)\right|_{x = 0}^{x=1} = -\dfrac{1}{2} \log^2(2) - \text{Li}_2(-1)}$

But $\text{Li}_2(-1) = \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \left(2^{1-2}-1\right) \zeta(2) = -\frac{1}{2} \zeta(2)$ .Thus

$\color{#D61F06}{\boxed{I=\dfrac{1}{2} \left( \zeta(2) - \log^2(2)\right)}}$