If $$\int \frac{x^2 - x + 1}{x^2 + 1}e^{\cot^{-1}x}dx = A(x)e^{\cot^{-1}x} + C$$, then $$A(x)$$ is equal to :

We are given that $$\int \frac{x^2 - x + 1}{x^2 + 1} e^{\cot^{-1}x} dx = A(x) e^{\cot^{-1}x} + C$$ and we need to find $$ A(x) $$.

To solve this, we use the substitution $$ t = \cot^{-1}x $$. Then, $$ x = \cot t $$. Differentiating both sides with respect to $$ t $$, we get $$ dx = -\csc^2 t dt $$.

Substituting into the integral:

$$\int \frac{(\cot t)^2 - \cot t + 1}{(\cot t)^2 + 1} e^{t} (-\csc^2 t) dt$$

We know that $$ \cot^2 t + 1 = \csc^2 t $$, so the denominator simplifies to $$ \csc^2 t $$. The numerator is $$ \cot^2 t - \cot t + 1 $$. Thus, the integral becomes:

$$\int \frac{\cot^2 t - \cot t + 1}{\csc^2 t} e^{t} (-\csc^2 t) dt$$

The $$ \csc^2 t $$ terms cancel:

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

Using the identity $$ \cot^2 t = \csc^2 t - 1 $$, we substitute:

$$-\int [(\csc^2 t - 1) - \cot t + 1] e^{t} dt = -\int (\csc^2 t - \cot t) e^{t} dt$$

This simplifies to:

$$-\int \csc^2 t e^{t} dt + \int \cot t e^{t} dt$$

Notice that $$ \csc^2 t - \cot t = -(\cot t - \csc^2 t) $$, so:

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

Consider the function $$ f(t) = e^t \cot t $$. Its derivative is:

$$f'(t) = e^t \cot t + e^t (-\csc^2 t) = e^t (\cot t - \csc^2 t)$$

Therefore, the integral $$ \int (\cot t - \csc^2 t) e^{t} dt = e^t \cot t + C $$.

Substituting back $$ t = \cot^{-1}x $$ and $$ \cot t = x $$:

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

So the integral is $$ x e^{\cot^{-1}x} + C $$.

Comparing with the given form $$ A(x) e^{\cot^{-1}x} + C $$, we have $$ A(x) e^{\cot^{-1}x} = x e^{\cot^{-1}x} $$, so $$ A(x) = x $$.

Verifying by differentiation: Differentiate $$ x e^{\cot^{-1}x} $$ using the product rule:

$$\frac{d}{dx}[x e^{\cot^{-1}x}] = e^{\cot^{-1}x} \cdot 1 + x \cdot e^{\cot^{-1}x} \cdot \left(-\frac{1}{1+x^2}\right) = e^{\cot^{-1}x} - \frac{x e^{\cot^{-1}x}}{1+x^2}$$

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

which matches the integrand.

Hence, the correct answer is Option B.