$\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x$ is equal to
Answer & explanation
Correct answer: option 1
$I=\int e^p\left(\sec ^2 p+\tan p\right) d p$ (If $p=\tan ^{-1} x \Rightarrow x=\tan p \Rightarrow d x=\sec ^2 p d p$)
$=e^{p} \tan p=x e^{\tan ^{-1} x}+c$
Hence (1) is the correct answer.