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Integrate the function
$\frac{e^{\tan ^{-1} x}}{1+x^2}$
$\frac{e^{\tan ^{-1} x}}{1+x^2}$
Solution:
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Verified Answer
Let $\tan ^{-1} x=t \Rightarrow \frac{1}{1+x^2} d x=d t$
$\therefore \int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int e^t d t=e^{\tan ^{-1} x}+\mathrm{C}$
$\therefore \int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int e^t d t=e^{\tan ^{-1} x}+\mathrm{C}$
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