Triple integral (check solution) The function given is
$f(x,y,z) = \displaystyle\frac{1}{(x+y+z+1)^2}$
$D = \{(x,y,z) \in \mathbb{R}^3 : x \geq 0, y \geq 0, z \geq 0, x+y+z \leq 1 \}$
It seems that the domain would be the under a plane that contains the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$; then if i consider the plane $XZ$ i have the set $D_{XZ}= \{(x,z) : z=1-x, 0 \leq x \leq 1 \}$ also $D_{XY}= \{(x,z) : y=1-x, 0 \leq x \leq 1\}$ and $D_{YZ}= \{(x,z) : z=1-y, 0 \leq y \leq 1\}$.
Then i could set $0\leq x\leq z-1$, $0\leq y\leq z-1$ and $0\leq z\leq 1$, and the integral would be given by:
$$\begin{align*} \displaystyle\int\displaystyle\int\displaystyle\int_D f(x,y,z)\,dx\,dy\,dz 
&= \displaystyle\int\displaystyle\int\displaystyle\int_D \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz 
\\&= \displaystyle\int_0^1\displaystyle\int_0^{1-z}\displaystyle\int_0^{1-z} \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz 
\\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{(1-z)+y+z+1)}- \displaystyle\frac{1}{(0)+y+z+1)}\bigg)\,dy\,dz
\\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{y+2}- \displaystyle\frac{1}{y+z+1}\bigg)\,dy\,dz
\\ &= -\displaystyle\int_0^1 (\log(y+2)-\log(y+z+1))|_{y=0}^{y=1-z}\,dz
\\ &= -\displaystyle\int_0^1 \{[(\log(3-z)-\log(2)] - [\log(2)-\log(z+1)]\}\,dz
\\ &= -\displaystyle\int_0^1 [(\log(3-z)+\log(z+1)-2\log(2)]\,dz
\\ &= 2\log(2)+\{[(3-z)\log(3-z) \\ & \;\;\;- (3-z)]|_{z=0}^{z=1} - [(z+1)\log(z+1)-(z+1)]|_{z=0}^{z=1}\}
\\ &= 2\log(2) - 3\log(3) +4
\end{align*} $$
Did i solved it correctly?

I don't think so.  I think the integral looks like
$$\int_0^1 dx \, \int_0^{1-x} dy \, \int_0^{1-x-y} dz \frac{1}{(1+x+y+z)^2}$$
which is equal to
$$-\int_0^1 dx \, \int_0^{1-x} dy \, \left [\frac{1}{2} - \frac{1}{1+x+y} \right ]$$
which is equal to
$$-\int_0^1 dx \, \left [\frac12 (1-x) - \log{2} + \log{(1+x)} \right ] $$
which in turn is equal to
$$-\frac14 + \log{2} - \left [ (1+x) \log{(1+x)} - (1+x)\right]_0^1 = \frac{3}{4}-\log{2}$$
This result was verified in Mathematica.
