We have,
$\begin{aligned}
& 5(f(x))^2=x f(x)+30 \Rightarrow 10 f(x) f^{\prime}(x)=x f^{\prime}(x)+f(x) \\
& \Rightarrow \quad(10 f(x)-x) f^{\prime}(x)=f(x)
\end{aligned}$
Now, $I=\int \frac{\left(3 x^3+\left(1-30 x^2\right) f(x)\right)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x$
$\begin{aligned}
& =\int \frac{3 x^3-30 x^2 f(x)+f(x)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x \\
& =\int \frac{3 x^2(x-10 f(x))+(10 f(x)-x) f^{\prime}(x)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x \\
& =\int \frac{-(10 f(x)-x)\left(3 x^2-f(x)\right)}{(10 f(x)-x)\left(x^3-f(x)\right)^3} d x \\
& =-\int \frac{3 x^2-f(x)}{\left(x^3-f(x)\right)^2} d x
\end{aligned}$
$\begin{aligned}
& \text { put } x^3-f(x)=t \\
& \Rightarrow \quad\left(3 x^2-f^{\prime}(x)\right) f(x)=d t \\
& \therefore \quad I=-\int \frac{d t}{t^2}=\frac{1}{t}+c \Rightarrow I=\frac{1}{x^3-f(x)}+c \\
& \therefore \quad A=1, B=1, D=-1, \\
& \therefore A+B+D=1+1-1=1
\end{aligned}$