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!
 A: Method 1
A trivial rescaling of the original integral ( substitute $x,y,z$ by $2x,2y,2z$ )
can bring it to the form of a type 1 Dirichlet integral 

Let $f$ be a function defined on
   $[0,1]$ and $\Delta \subset \mathbb{R}^d$ be the $d$-simplex $0 \le x_1, \ldots x_d$; $x_1+\cdots+x_d \le 1$, we have
$$\int_{\Delta} f(\sum_{i=1}^d x_i) \prod_{i=1}^{d} x_i^{\alpha_i-1} d^d x
=\frac{\prod_{i=i}^d \Gamma(\alpha_i)}{\Gamma(\sum_{i=1}^d \alpha_i)} \int_0^1 f(x) x^{\sum_{i=1}^d a_i)-1} dx$$

In this case $d = 3, f(x) = \frac{1}{(2x+1)^3}$ and $a_1 = a_2 = a_3 = 1$. 
It is then clear the integral is equal to
$$\frac{2^3\Gamma(1)^3}{\Gamma(3)}\int_0^1 \frac{x^{(3-1)}}{(2x+1)^3} dt
= 4 \left[ \frac{8x + 2(2x+1)^2\log(2x+1)+3}{16(2x+1)^2}\right]_0^1
= \frac{\log 3}{2} - \frac{4}{9}
$$
Method 2 
If one absolutely want to evaluate this as a multiple integral, there is a useful
trick to turn the integral as one over a hypercube $[0,1]^3$. Perform the substitution $x, y, z$ by $2x, 2y, 2z$ as before and introduce variables $\lambda, \mu, \nu$:
$$\begin{cases}
\lambda & = x + y + z\\
\lambda \mu & = y + z\\
\lambda \mu \nu & = z
\end{cases}
\quad\iff\quad
\begin{cases}
x &= \lambda (1-\mu)\\
y &= \lambda \mu (1-\nu)\\
z &= \lambda \mu \nu
\end{cases}$$
We have $$\begin{align}
dx \wedge dy \wedge dz &= d(x+y+z) \wedge d(y+z) \wedge dz = d\lambda \wedge \lambda d\mu \wedge \lambda\mu d\nu = \lambda^2\mu\,d\lambda \wedge d\mu \wedge d\nu
\end{align}$$
and the integral we want becomes
$$2^3 \int_{[0,1]^3} \frac{\lambda^2 \mu}{(2\lambda + 1)^3} d\lambda d\mu d\nu
= 4 \int_0^1 \frac{\lambda^2}{(2\lambda + 1)^3}  d\lambda$$
The same integral we got as a Dirichlet integral.
A: 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*}
