The Fubini trick

Write as double integral.
$$I=\int_0^1\int_0^1\frac{x}{(1+x^2)(1+xy)}\,dy\,dx.$$

Symmetry. Reversing $x\leftrightarrow y$:
$$2I=\int_0^1\int_0^1\frac{x+y}{(1+x^2)(1+y^2)}\,dx\,dy.$$

Separate.
$$2I=\left(\int_0^1\frac{x}{1+x^2}\,dx\right)\left(\int_0^1\frac{1}{1+y^2}\,dy\right)+\left(\int_0^1\frac{1}{1+x^2}\,dx\right)\left(\int_0^1\frac{y}{1+y^2}\,dy\right).$$
Both equal $\frac12\ln2\cdot\frac{\pi}{4}=\frac{\pi\ln2}{8}$. So $2I=\frac{\pi\ln2}{4}$, $I=\frac{\pi\ln2}{8}$.

Conclusion. $I=\frac{\pi\ln2}{8}$.