Question

$\\ \rm \int e^{x}\left(2021+\tan x+\tan ^{2} x\right) d x= \underline{ \qquad\qquad }+C$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \int e^{x}\left(2021+\tan x+\tan ^{2} x\right) d x= \underline{ \qquad\qquad }+c$

Let assume that

$\rm \int e^{x}\left(2021+\tan x+\tan ^{2} x\right) d x= g(x)+c$

So, it means we have to find the value of g(x)

So, Consider

$\rm \: \displaystyle\int\rm e^{x}\left(2021+\tan x+\tan ^{2} x\right)dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm e^{x}\left(2020 + 1+\tan x+\tan ^{2} x\right)dx$

can be re-arranged as

$\rm \: = \: \displaystyle\int\rm e^{x}\left(2020 +\tan x+[1 + \tan ^{2} x]\right)dx$

$\rm \: = \: \displaystyle\int\rm e^{x}\left(2020 +\tan x+\sec ^{2} x\right)dx$

$\rm \: = \: 2020\displaystyle\int\rm {e}^{x}dx \: + \: \displaystyle\int\rm {e}^{x}(tanx + {sec}^{2}x)dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x} \: [f(x) + f'(x)] \: dx \: = \: {e}^{x} \: f(x) + c \: \: }}$

and

$\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x} \: dx \: = \: {e}^{x} \: + \: c \: \: }}$

In second integral,

$\rm \: f(x) = tanx$

$\rm \: f'(x) = {sec}^{2}x$

So, using these results, we get

$\rm \: = \: 2020{e}^{x} \: + \: {e}^{x} \: tanx \: + \: c$

$\rm \: = \: {e}^{x} \:(2020 + \: tanx) \: + \: c$

Hence,

$\rm \: \displaystyle\int\rm e^{x}\left(2021+\tan x+\tan ^{2} x\right)dx = {e}^{x}(2020 + tanx) + c$

Now, it is given that

$\rm \int e^{x}\left(2021+\tan x+\tan ^{2} x\right) d x= g(x)+c$

So, on comparing two,

$\rm\implies \:\boxed{ \rm{ \:g(x) = {e}^{x} \: (2020 + tanx) \: \: }}$

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Remark :- Proof of the result

$\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x} \: [f(x) + f'(x)] \: dx \: = \: {e}^{x} \: f(x) + c \: \: }}$

Consider,

$\rm \: \:\displaystyle\int\rm {e}^{x} \: [f(x) + f'(x)] \: dx \:$

$\rm \: = \: \displaystyle\int\rm {e}^{x}f(x)dx \: + \: \displaystyle\int\rm {e}^{x}f'(x)dx$

On integrating first integral, by using Integration by parts,

$\rm \:=f(x)\displaystyle\int\rm {e}^{x}dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}f(x)\displaystyle\int\rm {e}^{x}dx \bigg]dx\: + \: \displaystyle\int\rm {e}^{x}f'(x)dx$

$\rm \: = \: f(x){e}^{x} - \displaystyle\int\rm f'(x){e}^{x}dx + \displaystyle\int\rm {e}^{x}f'(x)dx \: + \: c$

$\rm \: = \: {e}^{x} \: f(x) \: + \: c$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$