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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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