We have, $\frac{x^3}{(2 x-1)(x-1)^2}=\frac{x^3}{2 x^3-5 x^2+4 x-1}$


Now, consider
$$
\begin{aligned}
& \frac{1}{2}\left(\frac{5 x^2-4 x+1}{2 x^3-5 x^2+4 x-1}\right) \\
& \quad=\frac{1}{2}\left(\frac{B}{2 x-1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}\right) \\
& \frac{1}{2}\left(\frac{5 x^2-4 x+1}{2 x^3-5 x^2+4 x-1}\right) \\
& =\frac{1}{2}\left(\frac{B(x-1)^2+C(x-1)(2 x-1)+D(2 x-1)}{(2 x-1)(x-1)^2}\right) \\
& \Rightarrow 5 x^2-4 x+1 \\
& \quad=B(x-1)^2+C(x-1)(2 x-1)+D(2 x-1)
\end{aligned}
$$
Put $x=1$, we get
$$
D=2
$$
Put $x=1 / 2$, we get $B=+1$
Now put $x=0$, we get
$$
\begin{aligned}
1 & =B+C-D \\
\Rightarrow \quad 1 & =+1+C-2 \quad \Rightarrow \quad C=2
\end{aligned}
$$
Now, $\frac{1}{2}\left(\frac{5 x^2-4 x+1}{(2 x-1)(x-1)^2}\right)$
$$
\begin{gathered}
=\frac{1}{2}\left(\frac{+1}{2 x-1}+\frac{2}{x-1}+\frac{2}{(x-1)^2}\right) \\
=\frac{+\frac{1}{2}}{2 x-1}+\frac{\frac{2}{2}}{x-1}+\frac{1}{(x-1)^2}
\end{gathered}
$$
Put in Eq. (i), we get
$$
=\frac{1}{2}+\left(\frac{\frac{1}{2}}{2 x-1}+\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)
$$
By comparing $A=1 / 2, B=1 / 2, C=1, D=1$.
Now,
$$
\begin{aligned}
& 2 A-3 B+4 C+5 D \\
& =2 \times \frac{1}{2}-3 \times \frac{1}{2}+4 \times 1+5 \times 1 \\
& =1-\frac{3}{2}+4+5=10-\frac{3}{2}=\frac{17}{2}
\end{aligned}
$$