∫01x4(1−x)41+x2dx=

I=∫01x8−4x7+6x6−4x5+x41+x2dx =∫10(x6−4x5+5x4−4x2+4−41+x2)dx =17−23+1−43+4−π=227−π

∫01x4(1−x)41+x2dx=

I=∫01x8−4x7+6x6−4x5+x41+x2dx =∫10(x6−4x5+5x4−4x2+4−41+x2)dx =17−23+1−43+4−π=227−π

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

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$π - \frac{22}{7}$

$\frac{22}{7} - π$

$π + \frac{22}{7}$

answer is C.

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Detailed Solution

$I = \int_{0}^{1} \frac{x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4}}{1 + x^{2}} d x$

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

$= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - π = \frac{22}{7} - π$

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