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$\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} d x=$
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Let $\quad I=\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} d x$
Let $f(x)=\frac{x^{5}+x^{3}}{4-x^{2}} \Rightarrow f(-x)=\frac{-\left(x^{5}+x^{3}\right)}{4-x^{2}}$
$\therefore \quad \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{I}=0$
Let $f(x)=\frac{x^{5}+x^{3}}{4-x^{2}} \Rightarrow f(-x)=\frac{-\left(x^{5}+x^{3}\right)}{4-x^{2}}$
$\therefore \quad \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{I}=0$
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