The value(s) of $\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}\,dx$ is (are)

The value(s) of ∫₀¹(x⁴(1-x)⁴)/(1+x²) dx is (are) $I = \int_0^1 \dfrac{x^4(1-x)^4}{1+x^2}\,dx$ $\text{To find: the value of } I$ $1.\ \text{expand } x^4(1-x)^4$ $2.\ \text{divide by } 1+x^2$ $3.\ \text{integrate term by term on } [0,1]$ $(1-x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4$ $x^4(1-x)^4 = x^4 - 4x^5 + 6x^6 - 4x^7 + x^8$ $x^4(1-x)^4 = (1+x^2)\big(x^6 - 4x^5 + 5x^4 - 4x^2 + 4\big) - 4$ $\dfrac{x^4(1-x)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \dfrac{4}{1+x^2}$ $I = \left[\dfrac{x^7}{7} - \dfrac{4x^6}{6} + \dfrac{5x^5}{5} - \dfrac{4x^3}{3} + 4x\right]_0^1 - 4\!\!\int_0^1\!\!\dfrac{dx}{1+x^2}$ $= \dfrac{1}{7} - \dfrac{2}{3} + 1 - \dfrac{4}{3} + 4 - 4\cdot\dfrac{\pi}{4}$ $= \dfrac{1}{7} - 2 + 5 - \pi = \dfrac{1}{7} + 3 - \pi = \dfrac{22}{7} - \pi$ $\boxed{I = \dfrac{22}{7} - \pi \ \Rightarrow \ \text{option } (1)}$