Double integral where limits are the first quadrant

Evaluate the integral $$\iint\limits_D \frac{1}{(x+y+1)^3} \, dA$$ where $D$ is the first quadrant.

In this case, what would the limits of integration be? I'm having trouble moving to polar coordinates. Obviously the unit circle for the first quadrant is the area from $0$ to $\pi/2$, but I'm not sure how to break this up for $x$ and $y$.

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I think polar is not a good idea. Just do good old rectangular, $0$ to infinity twice. – André Nicolas Jul 28 '14 at 17:37
Ah, well, this just demonstrates my need for help even more. ;) – CODe Jul 28 '14 at 17:37
If we had $x^2+y^2$ stuff, I would go for polar. – André Nicolas Jul 28 '14 at 17:43

The quadrant is such a particular good area that you don't have to do any changes: $$\iint_D \frac{\mathrm d x\mathrm d y}{(1+x+y)^3} = \int_0^\infty\mathrm d y\int_0^\infty\frac{\mathrm d x}{(1+x+y)^3}$$ so first you'd compute $\int_0^\infty\frac{\mathrm d x}{(1+x+y)^3}$ as a function of $y$, and then integrate it.
Hint: Calculate $$\int_{x=0}^\infty\left(\int_{y=0}^\infty \frac{1}{(x+y+1)^3}\,dy\right)\,dx.$$ The inner integral is straightforward, just a power. Same for the second.
Maybe for the first you can let $u=x+y+1$, but it should not be necessary.