Explanation

(c)
Given, $\frac{x + 1}{(2x - 1)(3x + 1)} = \frac{A}{(2x - 1)} + \frac{B}{(3x + 1)}$
$\Rightarrow \ \ \ (x + 1) = A(3x + 1) + B(2x - 1)$
$\Rightarrow \ \ (x + 1) = x(3A + 2B) + A - B$
On equating the coefficient of $x$ and constant on both sides, we get
$3A + 2B = 1$ ...(i)
And $A - B = 1$ ...(ii)
On solving Eqs. (i) and (ii), we get
$A = \frac{3}{5},\ \ \ B = - \frac{2}{5}$
$\therefore 16A + 9B = 16\left( \frac{3}{5} \right) + 9\left( - \frac{2}{5} \right) = 6$