Solve integral $\int \int \int_A \frac{dxdydz}{(x+y+1)^2}$ where $A=\{(x,y,z):0<x<1, 0<x+y<1, 0<z(x+y+1)<1\}$

Question

I need to solve following integral $$ \int \int \int_A \frac{dxdydz}{(x+y+1)^2} $$ where $A=\{(x,y,z):0<x<1, 0<x+y<1, 0<z(x+y+1)<1\}$

If I do not take $z$ into consideration, I know how to solve it. I should use Fubini theorem here. Any advice would be extremely appreciated.

Answer 1

Maybe you could interchange the order?

$$\int\int\int_A\frac{dxdydz}{(x+y+1)^2}$$ $$=\ \int_0^1\int_{-x}^{1-x}\int_0^{\frac1{x+y+1}}\frac1{(x+y+1)^2}\ dzdydx$$ $$=\ \int_0^1\int_{-x}^{1-x}\frac1{(x+y+1)^3}\ dydx$$ $$=\ \int_0^1\left.\frac1{-2(x+y+1)^2}\right|_{y=-x}^{y=1-x}\,dx$$ $$=\ \int_0^1\frac38\ dx$$ $$=\frac38$$

Sorry if I make a mistake here, I have not been dealing with integral for quite a long time