Since,
$$
A(x)=\left|\begin{array}{ccc}
x+1 & 2 x+1 & 3 x+1 \\
2 x+1 & 3 x+1 & x+1 \\
3 x+1 & x+1 & 2 x+1
\end{array}\right|
$$
[applying $C_3 \rightarrow C_3-C_2$ and then $C_2 \rightarrow C_2-C_1$ ]
$\begin{aligned} & =\left|\begin{array}{ccc}x+1 & x & x \\ 2 x+1 & x & -2 x \\ 3 x+1 & -2 x & x\end{array}\right| \\ & \left.=x^2\left|\begin{array}{ccc}x+1 & 1 & 1 \\ 2 x+1 & 1 & -2 \\ 3 x+1 & -2 & 1\end{array}\right| \quad \text { [apply } C_2 \rightarrow C_2-C_3\right] \\ & =x^2\left|\begin{array}{ccc}x+1 & 0 & 1 \\ 2 x+1 & 3 & -2 \\ 3 x+1 & -3 & 1\end{array}\right| \quad \text { [apply } R_2 \rightarrow R_2+R_3 \text { ] }\end{aligned}$
$$
\begin{aligned}
& =x^2\left|\begin{array}{ccc}
x+1 & 0 & 1 \\
5 x+2 & 0 & -1 \\
3 x+1 & -3 & 1
\end{array}\right| \\
& =x^2(3)\{-x-1-5 x-2\} \\
& =3 x^2(-6 x-3)=-18 x^3-9 x^2 \\
& \therefore \quad \int_0^1 A(x) d x=\int_0^1\left(-18 x^3-9 x^2\right) d x \\
& =\left[\frac{-18 x^4}{4}-\frac{9 x^3}{3}\right]_0^1 \\
& =\left[\frac{-18}{4} x^4-\frac{9}{3} x^3\right]_0^1 \\
&
\end{aligned}
$$
[expand along $\mathrm{C}_2$ ]
$$
\begin{aligned}
& =-\frac{18}{4}-\frac{9}{3} \\
& =-\frac{18}{4}-3 \\
& =\frac{-18-12}{4} \\
& =-\frac{30}{4}=-\frac{15}{2}
\end{aligned}
$$