If $y = f(x^2)$, $\forall x \in \mathbb{R}$ is equal to
Step-by-Step Solution
Key Concept: Use substitution $x = u^3$ to simplify fractional powers, then integrate polynomial and logarithmic terms separately.
With $x = u^3$ and $dx = 6u^5du$, we have $\int\frac{6u^5du}{u} = \int 6u^4 du = \int(u^4 + u + 1 + \frac{1}{u})du$. After integration: $2u^5 - 3u^2 + 6u - 6\ln(u+1) + c = 2x^{5/3} - 3(x^{2/3}) + 6(x^{1/3}) - 6\ln(x^{1/3}+1) + c$. Using the point $(0,3)$: $3 = 0 - 4c^4 + c$ gives $c = 7$. Therefore $a = 2, b = -3, c = 6, d = -6$, and $20(2) + (-3) + 6 + (-6) = 37$.
Correct Answer: 37