Let f be a polynomial function such that $$f(x^{2}+1)=x^{4}+5x^{2}+2$$, for all $$x \in R$$. Then $$\int_{0}^{3}f(x)dx$$ is equal to

We need to find $$\int_0^3 f(x)dx$$ where $$f(x^2+1) = x^4 + 5x^2 + 2$$.

Let $$t = x^2 + 1$$, so $$x^2 = t - 1$$.

$$f(t) = (t-1)^2 + 5(t-1) + 2 = t^2 - 2t + 1 + 5t - 5 + 2 = t^2 + 3t - 2$$

So $$f(x) = x^2 + 3x - 2$$.

$$\int_0^3 (x^2 + 3x - 2)dx = \left[\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_0^3 = 9 + \frac{27}{2} - 6 = 3 + \frac{27}{2} = \frac{33}{2}$$

Therefore, the answer is Option 3: $$\frac{33}{2}$$.