A solution given by
akotronis :
Let $\displaystyle{S_{n}:=\sum_{k=1}^{n}(-1)^k\frac{\ln k}{k}}$. First observe that, by Dirichlet's criterion, the series converges, because $(-1)^k$ has bounded partial sums and $\displaystyle{\frac{\ln k}{k}}$ is eventually decreasing to $0$. Therefore $$\displaystyle\lim_{n\to+\infty}S_{2n}=\sum_{k=1}^{\infty}(-1)^k\frac{\ln k}{k}.$$ Now: \begin{align*}
S_{2n}&=\sum_{k=1}^{n}\frac{\ln2k}{2k}-\sum_{k=1}^{n}\frac{\ln(2k-1)}{2k-1}\\
&=\frac{\ln2}{2}\sum_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum_{k=1}^{n}\frac{\ln k}{k}-\left(\sum_{k=1}^{2n}\frac{\ln k}{k}-\sum_{k=1}^{n}\frac{\ln2k}{2k}\right)\\
&=\ln2\,H_{n}+\sum_{k=1}^{n}\frac{\ln k}{k}-\sum_{k=1}^{2n}\frac{\ln k}{k}\\
&=\ln2\,H_{n}-\sum_{k=1}^{n}\frac{\ln(n+k)}{n+k}\\
&=\ln2\,H_{n}-\sum_{k=1}^{n}\frac{\ln n+\ln(1+k/n)}{n+k}\\
&=\ln2\,H_{n}-\ln n\,(H_{2n}-H_{n})-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&=H_{n}\ln(2n)-H_{2n}\ln n-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&\stackrel{H_{n}=\ln n+\gamma+\mathcal O(1/n)}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\gamma\ln2+\mathcal O(1/n)-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&\longrightarrow\gamma\ln2-\int_{0}^{1}\frac{\ln(1+x)}{1+x}\,dx\\
&=\gamma\ln2-\frac{\ln^22}{2}\,.
\end{align*}