int_{-1}^1 frac{sin x-x^2}{3-|x|} d x= | vedclass

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(D) Let $I = \int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{\sin x}{3-|x|} d x - \int_{-1

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$7-18 \log \frac{3}{2}$

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(D) Let $I = \int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{\sin x}{3-|x|} d x - \int_{-1}^1 \frac{x^2}{3-|x|} d x$. Since $f(x) = \frac{\sin x}{3-|x|}$ is an odd function (because $\sin(-x) = -\sin x$ and $|-x| = |x|$),the integral $\int_{-1}^1 \frac{\sin x}{3-|x|} d x = 0$. Since $g(x) = \frac{x^2}{3-|x|}$ is an even function,$\int_{-1}^1 \frac{x^2}{3-|x|} d x = 2 \int_0^1 \frac{x^2}{3-x} d x$. Thus,$I = -2 \int_0^1 \frac{x^2}{3-x} d x = 2 \int_0^1 \frac{x^2}{x-3} d x$. Performing polynomial division: $\frac{x^2}{x-3} = x + 3 + \frac{9}{x-3}$. So,$I = 2 \int_0^1 (x + 3 + \frac{9}{x-3}) d x = 2 [\frac{x^2}{2} + 3x + 9 \ln|x-3|]_0^1$. Evaluating at limits: $I = 2 [(\frac{1}{2} + 3 + 9 \ln 2) - (0 + 0 + 9 \ln 3)] = 2 [\frac{7}{2} + 9 \ln(\frac{2}{3})] = 7 + 18 \ln(\frac{2}{3}) = 7 - 18 \ln(\frac{3}{2})$.

$\int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=$

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