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Question Number 20164 by tammi last updated on 23/Aug/17

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

Answered by ajfour last updated on 23/Aug/17

$$let{\tan }^{-1}x=t$$$$\Rightarrow \frac{dx}{1+x^2}=dt$$$$\Rightarrow \int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx=\int e^{t}dt=e^t+C$$$$=e^{{\tan }^{-1}x}+C.$$

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