Find $\displaystyle \int e^{\cot ^{-1} x}\left(\frac{1-x+x^{2}}{1+x^{2}}\right) \mathrm{d} x$. $\displaystyle \log \sqrt{3}$

$$\begin{aligned} & \text { Put } \cot ^{-1} x=t \therefore x=\cot t \text { and } \frac{1}{1+x^{2}} \mathrm{d} x=-\mathrm{d} t \\ & \begin{aligned} \therefore \int e^{\cot ^{-1} x}\left(\frac{1-x+x^{2}}{1+x^{2}}\right) \mathrm{d} x=-\int e^{t}\left(1-\cot t+\cot ^{2} t\right) \mathrm{d} t & =\int e^{t}\left(\cot t-\operatorname{cosec}^{2} t\right) \mathrm{d} t \\ & =e^{t} \cot t+c \\ & =x e^{\cot ^{-1} x}+c \end{aligned} \end{aligned} $$