Question

Evaluate: ${\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}dx$

Answer 1

Consider the given integral.

$I={\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}dx$

$I={\int }_0^1\frac{x^4{(1-x)}^2{(1-x)}^2}{1+x^2}dx$

$I={\int }_0^1\frac{x^4(1+x^2-2x)(1+x^2-2x)}{1+x^2}dx$

$I={\int }_0^1\frac{x^4(1+x^2-2x+x^2+x^4-2x^3-2x-2x^3+4x^2)}{1+x^2}dx$

$I={\int }_0^1\frac{x^4(1+6x^2+x^4-4x^3-4x)}{1+x^2}dx$

$I={\int }_0^1\frac{(x^4+6x^6+x^8-4x^7-4x^5)}{1+x^2}dx$

$I={\int }_0^1\frac{(x^8-4x^7+6x^6-4x^5+x^4)}{x^2+1}dx$

$I={\int }_0^1(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2})dx$

$I={[\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}-\frac{4x^3}{3}+4x-4{\tan }^{-1}x]}_0^1$

$I=[\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4{\tan }^{-1}\left(1\right)-\left[0\right]]$

$I=[\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-4{\tan }^{-1}\left(\tan \frac{\pi }{4}\right)]$

$I=[\frac{1}{7}-\frac{2}{3}-\frac{4}{3}+5-\frac{4\pi }{4}]$

$I=[\frac{1}{7}-2+5-\pi ]$

$I=[\frac{1}{7}-3-\pi ]$

$I=-(\frac{20}{7}+\pi )$

Hence, this is the answer.