The problem requires finding the value of the definite integral $\int_0^3 f(x)dx$, given a relationship involving the polynomial function $f$. The relationship is $f(x^2 + 1) = x^4 + 5x^2 + 2$.

Finding the Polynomial Function $f(x)$

To find the explicit form of $f(x)$, we use substitution. Let $y = x^2 + 1$. From this substitution, we have $x^2 = y - 1$. We can also express $x^4$ in terms of $y$: $x^4 = (x^2)^2 = (y - 1)^2$. Now, substitute $x^2$ and $x^4$ in the given equation $f(x^2 + 1) = x^4 + 5x^2 + 2$:

$f(y) = (y - 1)^2 + 5(y - 1) + 2$

Expand the terms:

$f(y) = (y^2 - 2y + 1) + (5y - 5) + 2$

Combine like terms to simplify the expression for $f(y)$:

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

Replacing $y$ with $x$, we get the function $f(x)$:

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

Calculating the Definite Integral $\int_0^3 f(x)dx$

Now we need to calculate the definite integral of $f(x)$ from $0$ to $3$. Substitute the expression for $f(x)$:

$ \int_0^3 (x^2 + 3x - 2)dx $

Find the antiderivative of $f(x)$: The antiderivative of $x^2$ is $\frac{x^3}{3}$. The antiderivative of $3x$ is $\frac{3x^2}{2}$. The antiderivative of $-2$ is $-2x$. So, the antiderivative is $\frac{x^3}{3} + \frac{3x^2}{2} - 2x$.

Evaluate the definite integral using the Fundamental Theorem of Calculus: $ \int_0^3 (x^2 + 3x - 2)dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_0^3 $

Substitute the upper limit ($x=3$) and the lower limit ($x=0$): $ \left( \frac{3^3}{3} + \frac{3(3^2)}{2} - 2(3) \right) - \left( \frac{0^3}{3} + \frac{3(0^2)}{2} - 2(0) \right) $

Calculate the values: $ \left( \frac{27}{3} + \frac{3(9)}{2} - 6 \right) - (0 + 0 - 0) $ $ \left( 9 + \frac{27}{2} - 6 \right) - 0 $ $ 3 + \frac{27}{2} $

Combine the terms to get the final result: $ \frac{6}{2} + \frac{27}{2} = \frac{33}{2} $

The calculated value of the integral is $\frac{33}{2}$.