The value of the integral $∫\left(e^x tan^{-1}x+\frac{e^x}{1+x^2}\right) dx $ is :

$e^x\, tan^{-1} x + C, $ where C is a constant

The correct answer is Option (1) → $e^x\, tan^{-1} x + C, $ where C is a constant

$\int e^x(\tan^{-1}x+\frac{1}{1+x^2})dx=e^x\tan^{-1}x+C$

$f(x)=\tan^{-1}x$

$f'(x)=\frac{1}{1+x^2}$

as $\int e^x\left(f(x)+f'(x)\right)dx=e^xf(x)+C$