Compute
$$I= \iiint \limits_S \frac{dx dy dz}{(1+x+y+z)^2}$$
where $S=\{x\ge 0, y\ge 0, z\ge 0, x+y+z\le 1\}$.
First, note that $0\le z \le 1-(x+y)$ where $(x, y)\in D=\{x\ge 0, y\ge 0, x+y \le 1\}$. Then
$$\begin{align*} I&=\iint \limits_D \left(\int_0^{1-(x+y)} \frac{1}{(1+x+y+z)^2}dz\right)dxdy\\&=\iint \limits_D \left(-\left.\frac{1}{1+x+y+z}\right|_0^{1-(x+y)}\right)dxdy\\&=\iint \limits_D \left(\frac{1}{1+x+y} - \frac{1}{2}\right)dxdy \end{align*}$$
Now, noting that $y\in[0, 1-x]$ and $x\in [0, 1]$ leads to
$$\begin{align*} I&= \int_0^1\left(\int_0^{1-x} \left(\frac{1}{1+x+y}-\frac{1}{2}\right) dy\right)dx\\&=\int_0^1\left(\left.\left(\ln{1+x+y}-\frac{1}{2}y\right)\right|_0^{1-x}\right)dx\\&=\int_0^1\left(\ln{2} -\ln(1+x) -\frac{1}{2}(1-x)\right)dx\\&=\left.\left((\ln{2})x -((1+x)\ln{x}-x) -(\frac{1}{2}x-\frac{1}{4}x^2)\right)\right|_0^1 \\&=\boxed{-\ln{2}+\frac{3}{4}} \end{align*}$$