Finding PDF from Multivariate Distribution and integrating out extra Variables Suppose X, Y and Z have the joint pdf
$f(x,y,z)=6(1+x+y+z)^{-4}$,  $x>0,y>0,z>0$
for $ U=x+y+z$
find pdf of $U$ so...
$v=y+z$,  $2<v<\infty$
$w=y$,   $0<w$
$z=w-v$, $y=w$
so $f_{u,v,w}(u,v,w)=6(1+u-v-w+w+w-v)^{-4}|J|=6(1+u-2v+w)^{-4}|J|$
where J=1 from the jacobian
to find the pdf of $U$ we must integrate out $v$ and $w$ which gets me 
$\int_1^\infty \int_2^\infty 6(1+u-2v+w)^{-4}dvdw=\int_1^\infty (-3+u+w)^{-3}dw$
$=\frac{1}{2}(u-2)^{-2}$ , $3<u<\infty$
so $f_u(u)=\frac{1}{2}(u-2)^{-2}$ , $3<u<\infty$

I think the calculation for $f_{U,V,W}(u,v,w)$ has a mistake. We have
\begin{align}
X &= U-V \\
Y &= W \\
Z &= V-W \\
& \\
\therefore\quad f_{U,V,W}(u,v,w) &= 6(1+u)^{-4}.
\end{align}
From the given values of $U,V,W,$ we deduce their ranges to be $0\lt W\lt V\lt U$. So the required integral to find $f_U(u)$ is:
\begin{align}
f_U(u) &= \int_{v=0}^u\int_{w=0}^v 6(1+u)^{-4}\;dwdv \\
&= \int_{v=0}^u 6v(1+u)^{-4}\;dv \\
&= \left[ 3v^2(1+u)^{-4} \right]_{v=0}^u \\
&= \dfrac{3u^2}{(1+u)^4} \qquad\qquad\text{for $u\gt 0$.}
\end{align}
