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Q18  Integrate the functions \frac{e ^{\tan ^{-1}x}}{1+ x^2 } 

Answers (1)

best_answer

Given,

\frac{e ^{\tan ^{-1}x}}{1+ x^2 }

Let's do the following substitution

 \\ tan^{-1}x = t \\ \implies \frac{1}{1+x^2}dx = dt

\therefore \int \frac{e ^{\tan ^{-1}x}}{1+ x^2 }dx = \int e ^{t}dt = e^t + C

= e^{tan^{-1}x} + C

Posted by

HARSH KANKARIA

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