int e^x(2021+tan x+tan^2 x) dx = . . . . . | vedclass

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(C) We know that $1 + 	an^2 x = \sec^2 x$. Substituting this into the integral,we get: $\int e^x(2021 + 	an x + 	an^2 x) dx = \int e^x(2020 + 1 + \

Vedclass · Accepted Answer

$(2020+	an x) e^x$

Vedclass · Answer

(C) We know that $1 + 	an^2 x = \sec^2 x$. Substituting this into the integral,we get: $\int e^x(2021 + 	an x + 	an^2 x) dx = \int e^x(2020 + 1 + 	an^2 x + 	an x) dx$ $= \int e^x(2020 + \sec^2 x + 	an x) dx$ $= \int e^x(2020 + 	an x) dx + \int e^x \sec^2 x dx$. Let $f(x) = 2020 + 	an x$. Then $f'(x) = \sec^2 x$. Using the standard integral formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + C$,we have: $\int e^x(2020 + 	an x + \sec^2 x) dx = e^x(2020 + 	an x) + C$. Thus,the correct option is $C$.

$\int e^x(2021+\tan x+\tan^2 x) dx = $ . . . . . . $+ C$.

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