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

Question

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Answer 1

Let $I = \int_{0}^{1} \frac{x ^{4} ( 1 - x ) ^{4}}{1 + x ^{2}} d x$
$= \int_{0}^{1} \frac{( x ^{4} - 1 ) ( 1 - x ) ^{4} + ( 1 - x ) ^{4}}{( 1 + x ^{2} )} d x$
$= \int_{0}^{1} (x^{2} - 1) (1 - x)^{4} d x + \int_{0}^{1} \frac{( 1 + x ^{2} - 2 x ) ^{2}}{( 1 + x ^{2} )} d x$
$= \int_{0}^{1} {(x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + \frac{4 x ^{2}}{( 1 + x ^{2} )}} d x$
$= \int_{0}^{1} ((x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + 4 \frac{4}{1 - x ^{2}}) d x$
$= \int_{0}^{1} (x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{1 + x ^{2}}) d x$
$= [\frac{x ^{7}}{7} - \frac{4 x ^{6}}{6} + \frac{5 x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x - 4 tan^{- 1} x]_{0}^{1}$
$= \frac{1}{7} - \frac{4}{6} + \frac{5}{5} - \frac{4}{3} + 4 - 4 (\frac{π}{4} - 0) = \frac{2 2}{7} - π$

Answer 2

Let

I = \int_{0}^{1} \frac{x ^{4} ( 1 - x ) ^{4}}{1 + x ^{2}} d x

= \int_{0}^{1} \frac{( x ^{4} - 1 ) ( 1 - x ) ^{4} + ( 1 - x ) ^{4}}{( 1 + x ^{2} )} d x

= \int_{0}^{1} (x^{2} - 1) (1 - x)^{4} d x + \int_{0}^{1} \frac{( 1 + x ^{2} - 2 x ) ^{2}}{( 1 + x ^{2} )} d x

= \int_{0}^{1} {(x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + \frac{4 x ^{2}}{( 1 + x ^{2} )}} d x

= \int_{0}^{1} ((x^{2} - 1) (1 - x)^{4} + (1 + x^{2}) - 4 x + 4 \frac{4}{1 - x ^{2}}) d x

= \int_{0}^{1} (x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{1 + x ^{2}}) d x

= [\frac{x ^{7}}{7} - \frac{4 x ^{6}}{6} + \frac{5 x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x - 4 tan^{- 1} x]_{0}^{1}

= \frac{1}{7} - \frac{4}{6} + \frac{5}{5} - \frac{4}{3} + 4 - 4 (\frac{π}{4} - 0) = \frac{2 2}{7} - π