(C) Let $t = \tan^{-1} x$.Then,differentiating with respect to $x$,we get $dt = \frac{1}{1 + x^2} dx$.Substituting these into the integral,we have:$\i

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(C) Let $t = \tan^{-1} x$.Then,differentiating with respect to $x$,we get $dt = \frac{1}{1 + x^2} dx$.Substituting these into the integral,we have:$\int \frac{e^{\tan^{-1} x}}{1 + x^2} dx = \int e^t dt$.The integral of $e^t$ is $e^t + c$.Substituting back $t = \tan^{-1} x$,we get $e^{\tan^{-1} x} + c$.