$\int\frac{e^{\tan^{-1}}x}{(1+x^2)}(1+x+x^2)dx$ is equal to :
Answer & explanation
Correct answer: option 3
Put $\tan^{-1}x=z$ i.e. x = tan z
$∴\int\frac{e^{\tan^{-1}}x}{1+x^2}(1+x+x^2)dx=\int e^z(1+\tan z+\tan^2z)dz=\int e^z(\tan z+sec^2z)dz$
$=\int e^z(\tan z+\frac{d}{dz}(\tan z))dz=e^z\tan z+C=xe^{\tan^{-1}x}+C$