I have to evaluate the following integral;

$\displaystyle \int_{-2}^{-1}\int_{-1}^{-2}\int_{-2}^{-1} x^2/(x^2+y^2+z^2) dx dy dz$

I sort of solved the first part (dx) with the integration formula:

$\displaystyle \int 1/(x^2+a^2) dx = 1/a *tan^{-1}(x/a)$

I did the following:

First I established $\displaystyle (y^2+z^2)=a^2$

$\displaystyle \int_{-2}^{-1} x^2/(x^2+a^2) dx= \int_{-2}^{-1} {1-a^2/(x^2+a^2)} dx= \int_{-2}^{-1} 1 dx -\int_{-2}^{-1} a^2/(x^2+a^2) dx=(-1-(-2)) -a^2*\int_{-2}^{-1} 1/(x^2+a^2) dx$

$\displaystyle =1 -a^2*(1/a *tan^{-1}(x/a))_{-2}^{-1}=1 -a^2*(1/a *tan^{-1}({-2}/a)-1/a *tan^{-1}({-1}/a))=1 -a*tan^{-1}({-2}/a)+ a*tan^{-1}({-1}/a))$

Then substitute a back to it's original values:

$\displaystyle 1 -\sqrt{(y^2+z^2)}*tan^{-1}({-2}/\sqrt{(y^2+z^2)})+ \sqrt{(y^2+z^2)}*tan^{-1}({-1}/\sqrt{(y^2+z^2)}))$

After this I have no idea what to do.

Any help would be appreciated but if possible I would like to know how to solve the entire problem.