$f(x) = \int e^{\tan^{-1}x}\left(1+x+x^2\right)d\left(\cot^{-1}x\right)$ is equal to:
Step-by-Step Solution
Key Concept: Recognize that d(cot⁻¹x) = -1/(1+x²)dx, transforming the integral into ∫e^(tan⁻¹x)(1+x+x²)·(-1/(1+x²))dx. Using substitution t = tan⁻¹x and simplifying (1+x+x²)/(1+x²) = 1 + x/(1+x²) leads to integration by parts or direct evaluation of -∫e^t·d(t) forms.
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