If ∫ x^2-x+1 x^2+1 e^cot ^-1 x d x=A(x) e^cot ^-1 x+C, then A(x) is equal to :

Let $\displaystyle \mathrm{I}=\int \frac{x^2-x+1}{x^2+1} \cdot e^{\cot ^{-1} x} d x$I=∫x2+1x2−x+1​⋅ecot−1xdx Put $x=\cot t \Rightarrow-\operatorname{cosec}^2 t d t=d x$x=cott⇒−cosec2tdt=dx Now, $1+\cot ^2 t=\operatorname{cosec}^2 t$1+cot2t=cosec2t $\therefore$∴ $\displaystyle \mathrm{I}=\int \frac{e^t\left(\cot ^2 t-\cot t+1\right)}{\left(1+\cot ^2 t\right)}\left(-\operatorname{cosec}^2 t\right) d t$I=∫(1+cot2t)et(cot2t−cott+1)​(−cosec2t)dt $$\begin{aligned} & =-\int e^t\left(\operatorname{cosec}{ }^2 t-\cot t\right) d t \\ & =\int e^t\left(\cot t-\operatorname{cosec}^2 t\right) d t \\ & =e^t \cot t+\mathrm{C} \\ & =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdot e^{\cot ^{-1} x}+\mathrm{C} \\ \Rightarrow \mathrm{A}(x) & =x \end{aligned}$$⇒A(x)​=−∫et(cosec2t−cott)dt=∫et(cott−cosec2t)dt=etcott+C=ecot−1x(x)+C≡A(x)⋅ecot−1x+C=x​