begin{aligned} & text{If } 5(f(x))^2 = x f(x) | vedclass

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(B) Given $5(f(x))^2 = x f(x) + 30$. Differentiating with respect to $x$, we get $10 f(x) f'(x) = x f'(x) + f(x)$, which implies $(10 f(x) - x) f'(x)

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(B) Given $5(f(x))^2 = x f(x) + 30$. Differentiating with respect to $x$, we get $10 f(x) f'(x) = x f'(x) + f(x)$, which implies $(10 f(x) - x) f'(x) = f(x)$.Let $I = \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx$.Substituting $f(x) = (10 f(x) - x) f'(x)$ into the numerator, we have:$I = \int \frac{3 x^3 - 30 x^2 f(x) + (10 f(x) - x) f'(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx$$I = \int \frac{3 x^2(x - 10 f(x)) + (10 f(x) - x) f'(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx$$I = \int \frac{-(10 f(x) - x)(3 x^2 - f'(x))}{(10 f(x) - x)(x^3 - f(x))^2} dx = -\int \frac{3 x^2 - f'(x)}{(x^3 - f(x))^2} dx$.Let $t = x^3 - f(x)$, then $dt = (3 x^2 - f'(x)) dx$.Thus, $I = -\int \frac{dt}{t^2} = \frac{1}{t} + C = \frac{1}{x^3 - f(x)} + C$.Comparing with $\frac{A}{B x^3 + D f(x)} + C$, we get $A = 1, B = 1, D = -1$.Therefore, $A + B + D = 1 + 1 - 1 = 1$.

$\begin{aligned} & \text{If } 5(f(x))^2 = x f(x) + 30 \text{ and } \\ & \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx \\ & = \frac{A}{B x^3 + D f(x)} + C, \text{ then } A + B + D = \end{aligned}$

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