$\int e^{x}\left( an^{-1}x+\dfrac{1}{1+x^2} ight)dx$ equals

Question

$\int e^{x}\left(	an^{-1}x+\dfrac{1}{1+x^2}ight)dx$ equals

Accepted Answer

A. $e^{x}	an^{-1}x+C$

1. Let $f(x)=	an^{-1}x$, then $f'(x)=\dfrac{1}{1+x^2}$.
2. The integrand is $e^{x}[f(x)+f'(x)]$.
3. Using $\int e^{x}[f(x)+f'(x)]\,dx=e^{x}f(x)+C$.
4. Result: $e^{x}	an^{-1}x+C$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.328_

Answer

$e^{x}\left(	an^{-1}x-\dfrac{1}{1+x^2}ight)+C$

Answer

$	an^{-1}x+C$

Answer

$\dfrac{e^{x}}{1+x^2}+C$

$\int e^{x}\left(\tan^{-1}x+\dfrac{1}{1+x^2}\right)dx$ equals

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