The correct option is (c) \(log\sqrt{2} – \frac{5}{16}\)
To elaborate: \(\int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{1-x-y} \frac{dz dy dx}{(1+x+y+z)^3} = \int_0^1 \int_0^{1-x}\Big[\frac{-1}{(2(1+x+y+z)^2}\Big]_{z=0}^{1-x-y} \,dy \,dx\)
\(\int_0^1 \int_0^{1-x}[\frac{-1}{8} + \frac{1}{(2(1+x+y)^2)}] \,dy \,dx = \int_0^1 \Big[\frac{-y}{8} – \frac{1}{2(1+x+y)}\Big]_{y=0}^{1-x} \,dx\)
\(\int_{x=0}^1 \Big[\frac{-(1-x)}{8} – \frac{1}{4} + \frac{1}{2(x+1)}\Big]dx = \int_{x=0}^1 \Big[\frac{-3}{8} + \frac{x}{8} + \frac{1}{2(x+1)}\Big]dx\)
\(\Big[\frac{-3x}{8} + \frac{x^2}{16} + \frac{1}{2} log(x+1)\Big]_{x=0}^1 = \frac{3}{8} + \frac{1}{16} + \frac{log2}{2} = log\sqrt{2} – \frac{5}{16}.\)