$\begin{gathered}
\frac{x+1}{(2 x-1)(3 x+1)}=\frac{A}{2 x-1}+\frac{B}{3 x+1} \\
x+1=A(3 x+1)+B(2 x-1)
\end{gathered}$
Put,
$x=\frac{1}{2} \text { in Eq. }$
$\begin{aligned}
& \frac{3}{2}=A\left(3 \cdot \frac{1}{2}+1\right)+B(0) \\
& \frac{3}{2}=A \cdot \frac{5}{2} \Rightarrow A=\frac{3}{5}
\end{aligned}$
Put,
$\begin{aligned}
& x=-\frac{1}{3} \text { in Eq. (i) } \\
& \frac{2}{3}=A(0)+B\left(2 \cdot\left(-\frac{1}{3}\right)-1\right) \\
& \frac{2}{3}=B\left(-\frac{5}{3}\right) \Rightarrow B=-\frac{2}{5}
\end{aligned}$
Consider
$16 A+9 B=16\left(\frac{3}{5}\right)+9\left(-\frac{2}{5}\right)=6$
Hence, option (3) is correct.