Integrating first w.r.t z we get

$\int\limits_0^{1}\int\limits_0^{1-x}\Bigg[\int\limits_0^{z=1-x-y}(1+x+y+z)^3dz\Bigg]dxdy$

$\int\limits_0^{1}\int\limits_0^{1-x}\dfrac 1{-2}\Bigg[\dfrac 1{(1+x+y+z)^2}\Bigg]^{1-x-y}_0dxdy$

$\dfrac 1{-2}\int\limits_0^{1}\int\limits_0^{1-x}\Bigg[\dfrac 1{(1+x+y+1-x-y)^2}-\dfrac 1{(1+x+y)^2}\Bigg]dxdy$

$\dfrac 1{-2}\int\limits_0^{1}\int\limits_0^{1-x}\Bigg[\dfrac 1{2^2}-\dfrac 1{(1+x+y)^2}\Bigg]dxdy$

Now integrating w.r.t x

$\dfrac 1{-2}\Bigg[\dfrac x4-\dfrac {x^2}8 +\dfrac x2 -\log(1-x)\Bigg]^1_0$

Putting upper and lower limits

We get

$=\dfrac 1{-2}\Bigg[\dfrac 14-\dfrac {1}8 +\dfrac 12 -\log 2\Bigg]$

$=\dfrac 1{-2}\Bigg[\dfrac 58 -\log 2\Bigg]$

$=\dfrac 1{2}\Bigg[\log 2 -\dfrac 58\Bigg]$