Answer
$${e^{{{\tan }^{ - 1}}x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}} dx \cr
& {\text{substitute }}u = {\tan ^{ - 1}}x, \cr
& du = \frac{1}{{1 + {x^2}}}dx \cr
& = \int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}} dx \cr
& = \int {{e^{{{\tan }^{ - 1}}x}}\frac{1}{{1 + {x^2}}}} dx \cr
& = \int {{e^u}du} \cr
& {\text{find the antiderivative }} \cr
& = {e^u} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {\tan ^{ - 1}}x, \cr
& = {e^{{{\tan }^{ - 1}}x}} + C \cr} $$
