Triple integral over a simplex with cube

Inner in $z$.
$$\int_0^{1-x-y}\frac{dz}{(x+y+z+1)^3}=\frac{1}{2(x+y+1)^2}-\frac18.$$

In $y$. $\frac12\int_0^{1-x}\frac{dy}{(x+y+1)^2}=\frac{1}{2}\left(\frac{1}{x+1}-\frac12\right)$, minus $\frac18(1-x)$.

In $x$.
$$I=\frac12\int_0^1\left(\frac{1}{x+1}-\frac12\right)dx-\frac18\cdot\frac12=\frac12\left(\ln2-\frac12\right)-\frac{1}{16}.$$

Result. $I=\frac12\ln2-\frac{5}{16}$.

Conclusion. $I=\frac12\ln2-\frac{5}{16}$.