We have,
$$
\begin{aligned}
I & =\int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x \\
& =-\frac{1}{p x^2+q x+r}+c \\
I & =\int \frac{2 x+3}{\left(x^2+3 x\right)\left(x^2+3 x+2\right)+1} d x
\end{aligned}
$$
Put $x^2+3 x=t \Rightarrow(2 x+3) d x=d t$
$$
\begin{array}{ll}
\therefore & I=\int \frac{d t}{t(t+2)+1} \\
& I=\int \frac{d t}{(t+1)^2}=\frac{-1}{t+1}+c \\
\Rightarrow & I=\frac{-1}{x^2+3 x+1}+c \\
\therefore & p=1, q=3, r=1 \\
\Rightarrow & \frac{3 p-q}{r}=\frac{3-3}{1}=0
\end{array}
$$