Question

Integrate the following question

Answer 1

$\implies\displaystyle \int {e^{x}}( {\tan}^{ - 1} (x) + \dfrac{1}{1 + {x}^{2} }) dx$

We know that :-

$\displaystyle \int {e^{x}}( f(x) + \frac{d(f(x))}{dx} ) \: dx = {e^{x}} f(x) + c$

Now we will use tye above Concept to integrate the given expression.

$\implies\displaystyle \int {e^{x}}( {\tan}^{ - 1} (x) + \dfrac{1}{1 + {x}^{2} }) dx$

$Here \: \: f(x) = {\tan}^{ - 1} (x)$

$\dfrac{d({\tan}^{ - 1} (x))}{dx} =\dfrac{1}{1 + {x}^{2} }$

Therefore :-

$\implies\displaystyle \int {e^{x}}( {\tan}^{ - 1} (x) + \dfrac{1}{1 + {x}^{2} }) dx = {e^{x}} {\tan}^{ - 1} (x) + c$

where c is constant.

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

Answer :

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

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

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