Limits of integral: $\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$ , where $D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$ What are the limits of the integral:
$$\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$$ where
$$ D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$$
I have previously done double integrals and was wondering if someone could help me derive the limits of this triple one.
Help is appreciated!

Hint: $(x,y,z)\in D$ if and only if $0<x<2$, $0<y<2-x$, and $0<z<2-x-y$. Then, $$\underset{D}{\iiint}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{(x+y+z+1)^3}=\int_{x=0}^{x=2}\left[\int_{y=0}^{y=2-x}\left(\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z\right)\mathrm{d}y\right]\mathrm{d}x.$$ Be careful: since the limits depend on other variables in some of the integrals on the right-hand side, the order of integration must not be changed! That this trick works is a consequence of Fubini's famous theorem.
Detailed calculations: $$\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z=\left[-\frac{1}{2(x+y+z+1)^2}\right]_{z=0}^{z=2-x-y}=-\frac{1}{18}+\frac{1}{2(x+y+1)^2}.$$
\begin{align*}\int_{y=0}^{y=2-x}\left(-\frac{1}{18}+\frac{1}{2(x+y+1)^2}\right)\mathrm{d}y=&\left[-\frac{y}{18}-\frac{1}{2(1+x+y)}\right]_{y=0}^{y=2-x}\\=&\,\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\end{align*}
\begin{align*}\int_{x=0}^{x=2}\left(\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\right)\mathrm{d}x=&\left[\frac{\dfrac{x^2}{2}-2x}{18}-\frac{x}{6}+\frac{\log (1+x)}{2}\right]_{x=0}^{x=2}\\=&\,-\frac{2}{18}-\frac{2}{6}+\frac{\log 3}{2}-0=-\frac{4}{9}+\frac{\log3}{2}.\end{align*}
