(A) We have $I = \int e^{x} \left( \tan^{-1} x + \frac{1}{1+x^{2}} \right) dx$. Consider $f(x) = \tan^{-1} x$. Then,the derivative is $f'(x) = \frac{1

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(A) We have $I = \int e^{x} \left( \tan^{-1} x + \frac{1}{1+x^{2}} \right) dx$. Consider $f(x) = \tan^{-1} x$. Then,the derivative is $f'(x) = \frac{1}{1+x^{2}}$. We know that the integral of the form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$. Comparing our integral with this standard form,we get $I = e^{x} \tan^{-1} x + C$.