Answer
$\dfrac{abc}{24}(a+4)$
Work Step by Step
$I=\int_{0}^{a}\int_0^{b(1-x/a)} \int_0^{c(1-x/a-y/b)} (1+x) dzdydx$
$=\int_{0}^{a}\int_0^{b(1-x/a)} (1+x) \times [z]_0^{c(1-x/a-y/b)}dydx$
Plug $t=1-\dfrac{x}{a} ; dt=\dfrac{-dx}{a}; dx=-adt$
$I=(ac) \int_{0}^{1} (1+a-at) \times [\dfrac{bt^2}{2}]dt=\dfrac{abc}{2} \int_{0}^{1} (t^2+at^2-at^3) dt$
or, $=\dfrac{abc}{2}[\dfrac{t^3}{3}+\dfrac{at^3}{3}-\dfrac{at^4}{4}]_0^1$
Hence, $I=(\dfrac{abc}{2}) \times (\dfrac{1^3}{3}+\dfrac{a(1)^3}{3}-\dfrac{a(1)^4}{4})=(\dfrac{abc}{2}) \times \dfrac{1}{3}+\dfrac{a(1)^3}{3}-\dfrac{a(1)^4}{4}=\dfrac{abc}{24}(a+4)$
