Math Archive (11)

$\begin{aligned} S & = \int_{-1}^1 \frac {x^4(1-x)^4}{1+x^2} dx & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-1}^1 \left(\frac {x^4(1-x)^4}{1+x^2} + \frac {x^4(1+x)^4}{1+x^2} \right) dx \\ & = \int_{-1}^1 \frac {x^4+6x^6+x^8}{1+x^2} \ dx \\ & = \int_{-1}^1 \frac {x^4+x^8}{1+x^2} \ dx + \color{#3D99F6} \int_{-1}^1 \frac {6x^6}{1+x^2} \ dx & \small \color{#3D99F6} \text{Since } \frac {6x^6}{1+x^2} \text{ is even,} \\ & = \int_{-1}^1 \left(x^6-x^4+2x^2-2+\frac 2{x^2+1} \right) dx + \color{#3D99F6} 12 \int_0^1 \frac {x^6}{1+x^2} \ dx \\ & = \frac {x^7}7 - \frac {x^5}5 + \frac {2x^3}3 - 2x + 2\tan^{-1} x \ \bigg|_{-1}^1 + 12 M \\ & = \frac 27 - \frac 25 + \frac 43 - 4 + \pi + 12 M \\ & \approx 0.3606 + 12 M \end{aligned}$

$\implies \lfloor S-12M\rfloor = \lfloor 0.3603 \rfloor = \boxed{0}$