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(B) Given that $f(x^2+1) = x^4 + 5x^2 + 2$. Let $t = x^2 + 1$, which implies $x^2 = t - 1$. Substituting this into the expression for $f$, we get: $f(

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$\frac{33}{2}$

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(B) Given that $f(x^2+1) = x^4 + 5x^2 + 2$. Let $t = x^2 + 1$, which implies $x^2 = t - 1$. Substituting this into the expression for $f$, we get: $f(t) = (t-1)^2 + 5(t-1) + 2$ $f(t) = (t^2 - 2t + 1) + 5t - 5 + 2$ $f(t) = t^2 + 3t - 2$. Now, we calculate the definite integral: $\int_{0}^{3} f(t) dt = \int_{0}^{3} (t^2 + 3t - 2) dt$ $= \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 2t \right]_{0}^{3}$ $= \left( \frac{3^3}{3} + \frac{3(3^2)}{2} - 2(3) \right) - (0)$ $= \left( \frac{27}{3} + \frac{27}{2} - 6 \right)$ $= 9 + 13.5 - 6 = 16.5 = \frac{33}{2}$.

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

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