The answer is an integer? Impossible!

$\begin{aligned} I & = \int_0^1 \frac {x^4(1-x)^4}{1+x^2} dx \\ & = \int_0^1 \frac {x^8-4x^7+6x^6-4x^5+x^4}{x^2+1} dx \\ & = \int_0^1 x^6 - 4x^5 + 5x^4 - 4x^2 + \frac 4{x^2+1} dx \\ & = \frac 17 - \frac 23 + 1 - \frac 43 + 4 - 4\tan^{-1} 1 \\ & = \frac {22}7- \pi \end{aligned}$

Therefore, $7(\pi - I) = \boxed {22}$ .

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Note: Since $I > 0$ , $\implies \frac {22}7 > \pi$ . In fact the integral is used to show that $\frac {22}7$ is only an approximation of $\pi$ .

use partial fraction techniques, then evaluate the problem

hence the answer 7[(10/3)+(1/105)-(1/5)] = 22

just expand the (1-x^4)...divide by x^4..then indefinite integrall.. and finally plug in the limits..

definite integral becomes 22/7 - pi

hence ANS is 22

7 (0.00126448926734962 + 3.14159265358979) =

7 (3.14285714285714000) = 22

Unbelievable.