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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

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