Question
Find $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$ 

Answer

We have $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$ 
Consider f(x) = tan-1 x, then f′(x) = $\frac{1}{1+x^{2}}$
Thus, the given integrand is of the form ex [f(x) + f′(x)].
Therefore, $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$ = ex tan-1 x + C

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