$\int\left(\frac{1+x+x^2}{1+x^2}\right)e^{\tan^{-1}x}dx=$
Answer & explanation
Correct answer: option 4
$\int\left(\frac{1+x+x^2}{1+x^2}\right)e^{\tan^{-1}x}dx$
$=\int\left(1+\frac{x}{1+x^2}\right)e^{\tan^{-1}x}dx$
this is of the form $\int e(f(x)+f'(x))dx$
as $\frac{dxe^{\tan^{-1}x}}{dx}=e^{\tan^{-1}x}+\frac{xe^{\tan^{-1}x}}{1+x^2}$
so $\int\left(1+\frac{x}{1+x^2}\right)e^{\tan^{-1}x}dx=xe^{\tan^{-1}x}+c$