$f(x) = \int e^{\tan^{-1}x}\left(1+x+x^2\right)d\left(\cot^{-1}x\right)$ is equal to:
Step-by-Step Solution
Key Concept: Recognizing that the derivative of $\tan^{-1}x$ appears in the integrand allows us to use substitution to reduce the problem to an integration by parts.
We rewrite the integrand using $\cot^{-1}x$ properties: $\int e^{\tan^{-1}x}(1+x+x^2)/(\cot^{-1}x) dx$ becomes $\int e^{\tan^{-1}x}(1+x^2)\frac{1}{1+x^2}dx$ after substitution $\tan^{-1}x = t$, so $dx = \frac{1}{1+x^2}dt$. This simplifies to $-\int e^t(\tan t + \sec^2 t)dt = -e^t \tan t + c = -xe^{\tan^{-1}x} + c$.
Correct Answer: 2,3