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(D) Let $I = \int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$.We can split this into two integrals: $I = \int_{-0.9}^{

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(D) Let $I = \int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$.We can split this into two integrals: $I = \int_{-0.9}^{0.9} [x^2] dx + \int_{-0.9}^{0.9} \log \left( \frac{2-x}{2+x} \right) dx$.For the first part,since $-0.9 For the second part,let $f(x) = \log \left( \frac{2-x}{2+x} \right)$.Then $f(-x) = \log \left( \frac{2-(-x)}{2+(-x)} \right) = \log \left( \frac{2+x}{2-x} \right) = \log \left( \left( \frac{2-x}{2+x} \right)^{-1} \right) = -\log \left( \frac{2-x}{2+x} \right) = -f(x)$.Since $f(x)$ is an odd function and the interval $[-0.9, 0.9]$ is symmetric about the origin,$\int_{-0.9}^{0.9} f(x) dx = 0$.Thus,$I = 0 + 0 = 0$.

If $[x]$ is the greatest integer $\leq x$,then the value of the integral $\int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$ is

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