Question

${\int }_0^{10}\frac{x^{10}}{{(10-x)}^{10}+x^{10}}dx$ is equal to

Answer 1

$I={\int }_0^{10}\frac{x^{10}}{{(10-x)}^{10}+x^{10}}\to \left(i\right)$

$\Rightarrow I={\int }_0^{10}\frac{{(10+0-x)}^{10}}{{[10-(10+0-x)]}^{10}+{[10+0-x]}^{10}}dx$

$\Rightarrow I={\int }_0^{10}\frac{{(10-x)}^{10}}{x^{10}+{[10-x]}^{10}}dx\to \left(ii\right)$

$\Rightarrow \left(i\right)+\left(ii\right)\to 2I={\int }_0^{10}\frac{{x^{10}+(10-x)}^{10}}{x^{10}+{[10-x]}^{10}}dx$

$\Rightarrow 2I={\int }_0^{10}dx$

$\Rightarrow 2I=10$

$\Rightarrow I=5.$

Hence, the answer is $5.$