Integral $\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x y) \ln (x y)} d x d y$

I think it is possible to avoid the Dirichlet eta function. Put\begin{equation*} I=\int_{0}^{1}\int_{0}^{1}\dfrac{1}{(1+xy)\ln(xy)}\,dxdy. \end{equation*} By symmetry \begin{equation*} I = 2\int_{0}^{1}\left(\int_{0}^{x}\dfrac{1}{(1+xy)\ln(xy)}\,dy\right)\, dx. \end{equation*} Via the substitution $y=\dfrac{z}{x}$ we get \begin{equation*} I = \int_{0}^{1}\left(\int_{0}^{x^2}\dfrac{1}{x(1+z)\ln(z)}\,dz\right)\,dx. \end{equation*} We have a double integral integrated over the domain $0<z<x^2, \, 0<x<1$. If we change the order of integration then we first have to integrate with respect to $x$ over $\sqrt{z}<x<1$ and then with respect to $z$ over $0<z<1.$ Thus \begin{gather*} I = 2\int_{0}^{1}\left(\int_{\sqrt{z}}^{1}\dfrac{1}{x(1+z)\ln(z)}\, dx\right)\, dz = 2\int_{0}^{1}\dfrac{1}{(1+z)\ln(z)}\left[\ln(x)\right]_{\sqrt{z}}^{1}\, dz =\\[2ex] \int_{0}^{1}\dfrac{-1}{1+z}\, dz = -\ln(2). \end{gather*}

$$I(a)=\int_0^1\int_0^1 \frac{(xy)^{a}}{(1+xy)\ln(xy)}dxdy\Rightarrow I'(a)=\int_0^1\int_0^1\frac{(xy)^a}{1+xy}dxdy$$ $$\overset{xy=t}=\int_0^1\frac1y\int_0^y\frac{t^a}{1+t}dtdy=\sum_{n=0}^\infty (-1)^n \int_0^1 \frac1{y}\int_0^y t^{a+n}dtdy$$ $$\require{cancel}=\sum_{n=0}^\infty (-1)^n \int_0^1 \cancel{\frac1{y}}\frac{y^{a+n+\cancel1}}{a+n+1}dy=\sum_{n=0}^\infty (-1)^n \frac{1}{(a+n+1)^2}=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(a+n)^2}$$ We are looking to find $I(0)$, but we also have that $\lim\limits_{a\to \infty}I(a)=0$ (see also the comments bellow why not $a\to -\infty$), thus by Newton-Leibniz formula: $$I(0)=-(I(\infty)-I(0))=-\int^{\infty}_0 I'(a)da=-\sum_{n=1}^\infty(-1)^{n-1} \int^{\infty}_0 \frac{1}{(a+n)^2}da$$ $$=-\sum_{n=1}^\infty(-1)^{n-1}\left(-\frac{1}{a+n}\right)\bigg|^{\infty}_0 =-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=-\ln 2 $$