int x^{2020}(tan^{-1} x + cot^{-1} x) dx = | vedclass

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(B) We know that for all $x \in \mathbb{R}$,$	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.Substituting this into the integral,we get:$I = \int x^{2020}

Vedclass · Accepted Answer

$\frac{x^{2021}}{2021}(	an^{-1} x + \cot^{-1} x) + C$

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(B) We know that for all $x \in \mathbb{R}$,$	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.Substituting this into the integral,we get:$I = \int x^{2020} \left(\frac{\pi}{2}ight) dx$$I = \frac{\pi}{2} \int x^{2020} dx$Using the power rule for integration,$\int x^n dx = \frac{x^{n+1}}{n+1} + C$:$I = \frac{\pi}{2} \cdot \frac{x^{2021}}{2021} + C$Since $\frac{\pi}{2} = 	an^{-1} x + \cot^{-1} x$,we can write:$I = \frac{x^{2021}}{2021} (	an^{-1} x + \cot^{-1} x) + C$Thus,option $B$ is correct.

$\int x^{2020}(\tan^{-1} x + \cot^{-1} x) dx =$

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