Mathematics

# Solve $\displaystyle \int \dfrac { { e }^{ \tan ^{ -1 }{ x } } }{ 1+{ x }^{ 2 } }dx$

##### SOLUTION
$Let, I =\displaystyle \int \dfrac{e^{\tan^{-1}} x}{1 + x^2} dx$

Put                  $\tan^{-1} x = t$

$\implies \dfrac{1}{1 + x^2} dx = dt$

$\therefore I = \int e^{t} dt$

$I = e^{t} + c = e^{\tan^{-1} x} + c$ (Ans)

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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