Find $\int e^{\cot^{-1} x} \left( \frac{1-x+x^2}{1+x^2} \right) dx$.

$x e^{\cot^{-1} x} + C$

The correct answer is Option (2) → $x e^{\cot^{-1} x} + C$

$\int e^{\cot^{-1} x} \left( \frac{1-x+x^2}{1+x^2} \right) dx$

Let $\cot^{-1} x = t ⇒\frac{-1}{1+x^2} dx = dt$

$\int e^{\cot^{-1} x} \left( \frac{1-x+x^2}{1+x^2} \right) dx = \int -e^t (1 - \cot t + \cot^2 t) dt$

$= \int -e^t (1 - \cot t + \text{cosec}^2 t - 1) dt$

$= \int -e^t (\text{cosec}^2 t - \cot t) dt$

$= \int e^t (\cot t - \text{cosec}^2 t) dt$

If $f(t) = \cot t$, then $f'(t) = -\text{cosec}^2 t$

$= \int e^t (f(t) + f'(t)) dt$

$= e^t (f(t)) + C$

$= e^t \cot t + C$

$= e^{\cot^{-1} x} \cot(\cot^{-1} x) + C$

$= x \cdot e^{\cot^{-1} x} + C$