The value(s) of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

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Aicharya · Accepted Answer

Let $I=\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ $ \begin{aligned} & =\int_0^1 \frac{\left(x^4-1ight)(1-x)^4+(1-x)^4}{\left(1+x^2ight)} d x \ & =\int_0^1\left(x^2-1ight)(1-x)^4 d x \ & +\int_0^1 \frac{\left(1+x^2-2 xight)^2}{\left(1+x^2ight)} d x \ & =\int_0^1\left\{\left(x^2-1ight)(1-x)^4+\left(1+x^2ight)-4 xight. \ & \left.+\frac{4 x^2}{\left(1+x^2ight)}ight\} d x \ & =\int_0^1\left(\left(x^2-1ight)(1-x)^4+\left(1+x^2ight)-4 xight. \ & \left.+4-\frac{4}{1-x^2}ight) d x \ & =\int_0^1\left(x^6-4 x^5+5 x^4-4 x^2ight. \ & \left.+4-\frac{4}{1+x^2}ight) d x \ & =\left[\frac{x^7}{7}-\frac{4 x^6}{6}+\frac{5 x^5}{5}ight. \ & \end{aligned} $ $ \begin{aligned} & \left.\quad-\frac{4 x^3}{3}+4 x-4 	an ^{-1} xight]_0^1 \ = & \frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi}{4}-0ight) \ = & \frac{22}{7}-\pi \end{aligned} $

The value(s) of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

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