$\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} d x=$

Question

Aicharya · Accepted Answer

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

$\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} d x=$

Solution