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
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