The value of the integral int_{-1}^{1} left{ | vedclass

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(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$.We can split the integral into two parts:$I

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$2(1-e^{-1})$

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(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$.We can split the integral into two parts:$I = \int_{-1}^{1} \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} dx + \int_{-1}^{1} \frac{1}{e^{|x|}} dx$.Let $f(x) = \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)}$. Since $f(-x) = \frac{(-x)^{2013}}{e^{|-x|}((-x)^{2}+\cos(-x))} = \frac{-x^{2013}}{e^{|x|}(x^{2}+\cos x)} = -f(x)$, the function $f(x)$ is an odd function. Therefore, $\int_{-1}^{1} f(x) dx = 0$.Let $g(x) = \frac{1}{e^{|x|}} = e^{-|x|}$. Since $g(-x) = e^{-|-x|} = e^{-|x|} = g(x)$, the function $g(x)$ is an even function. Therefore, $\int_{-1}^{1} g(x) dx = 2 \int_{0}^{1} e^{-x} dx$.Calculating the integral: $2 \int_{0}^{1} e^{-x} dx = 2 [-e^{-x}]_{0}^{1} = 2 (-e^{-1} - (-e^{0})) = 2(1 - e^{-1})$.Thus, $I = 0 + 2(1 - e^{-1}) = 2(1 - e^{-1})$.

The value of the integral $\int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$ is equal to

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