$$\int \frac{(x^2+1)e^x}{(x+1)^2} dx = f(x)e^x + C$$, where $$C$$ is a constant, then $$\frac{d^3f}{dx^3}$$ at $$x = 1$$ is equal to

We need to find $$f(x)$$ such that $$\int \frac{(x^2+1)e^x}{(x+1)^2} dx = f(x)e^x + C$$.

First, simplify the integrand by writing $$x^2 + 1 = (x+1)^2 - 2x = (x+1)^2 - 2(x+1) + 2$$, which gives $$\frac{x^2+1}{(x+1)^2} = 1 - \frac{2}{x+1} + \frac{2}{(x+1)^2}$$.

Next, we use the identity for integrals of the form $$\int [g(x) + g'(x)]e^x dx = g(x)e^x + C$$, so it is necessary to express $$\frac{x^2+1}{(x+1)^2}$$ in the form $$g(x) + g'(x)$$. To this end, let $$g(x) = 1 - \frac{2}{x+1} + \frac{a}{(x+1)^2}$$ for some constant $$a$$. Differentiating gives $$g'(x) = \frac{2}{(x+1)^2} - \frac{2a}{(x+1)^3}$$, and therefore $$g(x) + g'(x) = 1 - \frac{2}{x+1} + \frac{a+2}{(x+1)^2} - \frac{2a}{(x+1)^3}$$. Comparing this with $$1 - \frac{2}{x+1} + \frac{2}{(x+1)^2}$$ shows that $$a + 2 = 2$$ and $$-2a = 0$$, so $$a = 0$$.

Hence $$g(x) = 1 - \frac{2}{x+1} = \frac{x-1}{x+1}$$. Checking, we find $$g'(x) = \frac{(x+1) - (x-1)}{(x+1)^2} = \frac{2}{(x+1)^2}$$, and then $$g(x) + g'(x) = \frac{x-1}{x+1} + \frac{2}{(x+1)^2} = \frac{(x-1)(x+1) + 2}{(x+1)^2} = \frac{x^2 + 1}{(x+1)^2}$$, as required. Therefore, $$f(x) = \frac{x-1}{x+1}$$.

To find the third derivative of $$f$$ at $$x = 1$$, note that $$f(x) = \frac{x-1}{x+1} = 1 - \frac{2}{x+1}$$ gives $$f'(x) = \frac{2}{(x+1)^2}$$, $$f''(x) = \frac{-4}{(x+1)^3}$$, and $$f'''(x) = \frac{12}{(x+1)^4}$$. Substituting $$x = 1$$ yields $$f'''(1) = \frac{12}{2^4} = \frac{12}{16} = \frac{3}{4}$$.

The answer is Option A: $$\frac{3}{4}$$.