\intx^{2020}(\Tan^{-1} x+\Cot^{-1} x) dx=

Correct Answer is : x20212021 (Tan−1x+Cot−1x)+c

{x^{2021}/2020} (\Tan^{-1} x+\Cot^{-1} x)+c

{x^{2021}/2021} (\Tan^{-1} x+\Cot^{-1} x)+c

AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 23 of 160

Marks: +1, -0

∫x2020(Tan−1x+Cot−1x) dx=

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