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Q.
The value(s ) of $ \int\limits^1_0 \frac { x^4 ( 1 - x )^4 }{ ( 1 + x^2 ) } \, dx $ is are
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Solution:
Let $I =\int\limits_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x$
$=\int\limits_{0}^{1} \frac{\left(x^{4}-1\right)(1-x)^{4}+(1-x)^{4}}{\left(1+x^{2}\right)} d x $
$=\int\limits_{0}^{1}\left(x^{2}-1\right)(1-x)^{4} d x+\int\limits_{0}^{1} \frac{\left(1+x^{2}-2 x\right)^{2}}{\left(1+x^{2}\right)} d x$
$=\int\limits_{0}^{1}\left\{\left(x^{2}-1\right)(1-x)^{4}+\left(1+x^{2}\right)-4 x+\frac{4 x^{2}}{\left(1+x^{2}\right)}\right\} d x $
$=\int\limits_{0}^{1}\left(\left(x^{2}-1\right)(1-x)^{4}+\left(1+x^{2}\right)-4 x+4-\frac{4}{1+x^{2}}\right) d x$
$=\int\limits_{0}^{1}\left(x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{1+x^{2}}\right) d x$
$=\left[\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\right]_{0}^{1} $
$=\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi}{4}-0\right)=\frac{22}{7}-\pi$