$\cfrac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}$
$\cfrac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}$
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NCERT & Exemplar
: Let $I = \int {\cfrac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx}$
Put ${\tan ^{ - 1}}x = t$ $\Rightarrow$ $\cfrac{1}{{1 + {x^2}}}dx = dt$
$\therefore$ $I = \int {{e^t}dt} = {e^t} + C = {e^{{{\tan }^{ - 1}}x}} + C$
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