The value of int_{ - 1}^1 {frac{{sin x - {x^2 | vedclass

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(C) Let $I = \int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}} \,dx$. We can split the integral as: $I = \int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \

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$2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$

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(C) Let $I = \int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}} \,dx$. We can split the integral as: $I = \int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx - \int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$. Let $f(x) = \frac{{\sin x}}{{3 - |x|}}$ and $g(x) = \frac{{{x^2}}}{{3 - |x|}}$. For $f(x)$,$f(-x) = \frac{{\sin(-x)}}{{3 - |-x|}} = \frac{{-\sin x}}{{3 - |x|}} = -f(x)$,so $f(x)$ is an odd function. Thus,$\int_{ - 1}^1 {\frac{{\sin x}}{{3 - |x|}}} \,dx = 0$. For $g(x)$,$g(-x) = \frac{{(-x)^2}}{{3 - |-x|}} = \frac{{{x^2}}}{{3 - |x|}} = g(x)$,so $g(x)$ is an even function. Thus,$\int_{ - 1}^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx$. Substituting these back into $I$: $I = 0 - 2\int_0^1 {\frac{{{x^2}}}{{3 - |x|}}} \,dx = 2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$.

The value of $\int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $ is

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