Answer: $\mathbf C$
Explanation:
Using partial differentiation:
$\frac{x}{(x-1)(2x+3)} = \frac{\mathrm A}{x-1} + \frac{\mathrm B}{(2x+3)}$
$\Rightarrow \mathrm x = \mathrm A(2x+3) + \mathrm B(x-1)$
When $x = 1$, then $1 = 5\mathrm A \Rightarrow \mathrm A = \frac{1}{5}$
when $x = -\frac{3}{2}, \text {then}\; -\frac{3}{2} = -\frac{5\mathrm B}{2} \Rightarrow \mathrm B = \frac{3}{5}$
So, $f(x) = \frac{x}{(x-1)(2x+3)} = \frac{1}{(x-1)}+\frac{3}{5(3x+3)}$
Now, $f(x) = \frac{1}{5(x-1)} + \frac{3}{5(2x+3)}\\ f'(x) =\frac{-1}{5(x-1)^2} - \frac{6}{5(2x+3)^2}\\ f''(x) = \frac{2}{5(x-1)^3}+ + \frac{24}{5(2x+3)^3}\\ f'''(x) = \frac{-6}{5(x-1)^4} - \frac{72\times2}{5(2x+3)^4} \\f''''(x) = \frac{24}{5(x-1)^5} + \frac{72\times2\times8}{5(2x+3)^5} \\f''''(x) = \frac{24}{5}\bigg[\frac{1}{(x-1)^5} + \frac{48}{5(2x+3)^5}\bigg ]$
$\therefore \mathrm {\mathbf C}$ is the correct option.