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$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=$
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Verified Answer
The correct answer is:
$e^{\tan ^{-1} x}+c$
Putting $\quad t=\tan ^{-1} x \Rightarrow d t=\frac{1}{1+x^2} d x$, we get
$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int e^t d t=e^t+c=e^{\tan ^{-1} x}+c .$
$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int e^t d t=e^t+c=e^{\tan ^{-1} x}+c .$
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