Mathematics

Evaluate: $$\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ \left( 1-x \right)  }^{ 4 } }{ 1+{ x }^{ 2 } }  } dx$$


SOLUTION

Consider the given integral.

$$I=\int_{0}^{1}{\dfrac{{{x}^{4}}{{\left( 1-x \right)}^{4}}}{1+{{x}^{2}}}}dx$$

$$ I=\int_{0}^{1}{\dfrac{{{x}^{4}}{{\left( 1-x \right)}^{2}}{{\left( 1-x \right)}^{2}}}{1+{{x}^{2}}}}dx $$

$$ I=\int_{0}^{1}{\dfrac{{{x}^{4}}\left( 1+{{x}^{2}}-2x \right)\left( 1+{{x}^{2}}-2x \right)}{1+{{x}^{2}}}}dx $$

$$ I=\int_{0}^{1}{\dfrac{{{x}^{4}}\left( 1+{{x}^{2}}-2x+{{x}^{2}}+{{x}^{4}}-2{{x}^{3}}-2x-2{{x}^{3}}+4{{x}^{2}} \right)}{1+{{x}^{2}}}}dx $$

$$ I=\int_{0}^{1}{\dfrac{{{x}^{4}}\left( 1+6{{x}^{2}}+{{x}^{4}}-4{{x}^{3}}-4x \right)}{1+{{x}^{2}}}}dx $$

$$ I=\int_{0}^{1}{\dfrac{\left( {{x}^{4}}+6{{x}^{6}}+{{x}^{8}}-4{{x}^{7}}-4{{x}^{5}} \right)}{1+{{x}^{2}}}}dx $$

$$ I=\int_{0}^{1}{\dfrac{\left( {{x}^{8}}-4{{x}^{7}}+6{{x}^{6}}-4{{x}^{5}}+{{x}^{4}} \right)}{{{x}^{2}}+1}}dx $$

$$ I=\int_{0}^{1}{\left( {{x}^{6}}-4{{x}^{5}}+5{{x}^{4}}-4{{x}^{2}}+4-\dfrac{4}{1+{{x}^{2}}} \right)}dx $$

$$ I=\left[ \dfrac{{{x}^{7}}}{7}-\dfrac{4{{x}^{6}}}{6}+\dfrac{5{{x}^{5}}}{5}-\dfrac{4{{x}^{3}}}{3}+4x-4{{\tan }^{-1}}x \right]_{0}^{1} $$

$$ I=\left[ \dfrac{1}{7}-\dfrac{4}{6}+\dfrac{5}{5}-\dfrac{4}{3}+4-4{{\tan }^{-1}}\left( 1 \right)-\left[ 0 \right] \right] $$

$$ I=\left[ \dfrac{1}{7}-\dfrac{2}{3}+1-\dfrac{4}{3}+4-4{{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right) \right] $$

$$ I=\left[ \dfrac{1}{7}-\dfrac{2}{3}-\dfrac{4}{3}+5-\dfrac{4\pi }{4} \right] $$

$$ I=\left[ \dfrac{1}{7}-2+5-\pi  \right] $$

$$ I=\left[ \dfrac{1}{7}-3-\pi  \right] $$

$$ I=-\left( \dfrac{20}{7}+\pi  \right) $$

 

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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