Evaluate $\int e^{\tan^{-1}x}(1+x+x^2)d(\cot^{-1}x)$

Question

Integrate: $$\displaystyle \int e^{\tan^{-1}x}(1+x+x^2)d(\cot^{-1}x)$$

How to solve it? Integration with respect to $\cot^{-1}x$. I have started to change this integrand to integrate with respect to other variable, $\cot^{-1}x=t$

Answer 1

Let $t = \tan^{-1} x \Rightarrow x = \tan t$, and use : $\cot^{-1}x + \tan^{-1}x = \dfrac{\pi}{2}$ we have: $d(\cot^{-1}x) = -d(\tan^{-1}x) = -dt \Rightarrow I = -\displaystyle \int e^{t}(1+\tan t + \tan^2t)dt$.

We have: $\displaystyle \int e^{t}\tan t dt = \displaystyle \int \tan td(e^t) = e^t\tan t - \displaystyle \int e^t(1+\tan^2 t)dt \Rightarrow I = -e^t\tan t = -xe^{\arctan x}$