If $I=\int_{-1}^1([x]^2+\log(\frac{2+x}{2-x}))dx$ where [x] denotes the greatest integer ≤ x, then I equals:

-2 -1 0 1

0

[x]2 = 0  x ∈ (-1,1) and $I=\int\limits_{-1}^1\underset{odd\,function}{\underbrace{\log(\frac{2+x}{2-x})}}$