Mathematics

$$\displaystyle \int e^{\tan ^{-1}x}(1+x+x^{2})  d(\cot ^{-1}x)$$ is equal to


ANSWER

$$\displaystyle -xe ^{\displaystyle \tan ^{-1}x}+c$$


SOLUTION
$$I=\int { { e }^{ \tan ^{ -1 }{ x }  }(1+x+{ x }^{ 2 }) d(\cot ^{ -1 }{ x } ) }$$
$$\because d(\cot ^{ -1 }{ x } )=\frac { -1 }{ 1+{ x }^{ 2 } } dx$$
$$\Rightarrow$$ $$I=\int { { e }^{ \tan ^{ -1 }{ x }  }(1+x+{ x }^{ 2 })(\frac { -1 }{ 1+{ x }^{ 2 } } )dx }$$
$$\Rightarrow$$ $$I=-\int { { e }^{ \tan ^{ -1 }{ x }  }(1+x(1+{ x }))(\frac { 1 }{ 1+{ x }^{ 2 } } )dx }$$
$$\Rightarrow$$ $$I=-\int { { e }^{ \tan ^{ -1 }{ x }  }(1+\frac { x }{ { 1+x }^{ 2 } } )dx }$$
$$\Rightarrow$$ $$I=-\int { d(x{ e }^{ \tan ^{ -1 }{ x }  }) }$$
$$\Rightarrow$$ $$I=-x{ e }^{ \tan ^{ -1 }{ x }  }+C$$    (where $$C$$ is a constant)
Hence, option (C) is correct.
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Single Correct Hard Published on 17th 09, 2020
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