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

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

$\int e^{\tan^{-1}x}(1+x+x^2)d(\cot^{-1}x)=$

Options:

$e^{\tan^{-1}x}+c$

$-e^{\tan^{-1}x}+c$

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

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

Correct Answer:

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

Explanation:

$I=-\int e^{\tan^{-1}x}(1+x+x^2).\frac{1}{1+x^2}dx$

$=-\int e^{\tan^{-1}x}dx-\int\frac{e^{\tan^{-1}x}}{1+x^2}.dx$

$=-\int e^{\tan^{-1}x}dx-\{e^{\tan^{-1}x}.x-\int e^{\tan^{-1}x}.1\,dx\}+c$

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