Question
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = } $
अत: $\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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