We have,$\frac{x^3}{(2x-1)(x-1{)}^2}=A+\frac{B}{2x-1}+\frac{C}{\left(x-1\right)}+\frac{D}{{\left(x-1\right)}^2}$

Now,

${\text{D}}^r\text{=}(2x-1)(x-1{)}^2$

     $=(2x-1)(x^2-2x+1)$

       $=2x^3-5x^2+4x-1$

Now,

$\frac{x^3}{(2x-1)(x-1{)}^2}=\frac{1}{2}\left\{\frac{2x^3}{(2x-1)(x-1{)}^2}\right\}$

                                $=\frac{1}{2}\left\{\frac{2x^3-5x^2+4x-1+\left(5x^2-4x+1\right)}{(2x-1)(x-1{)}^2}\right\}$

 $=\frac{1}{2}\left\{1+\frac{\left(5x^2-4x+1\right)}{(2x-1)(x-1{)}^2}\right\}$

$=\frac{1}{2}+\frac{1}{2}\left\{\frac{\left(5x^2-4x+1\right)}{(2x-1)(x-1{)}^2}\right\} ...\left(i\right)$

Now,

$\frac{\left(5x^2-4x+1\right)}{\left(2x-1\right){\left(x-1\right)}^2}=\frac{B}{2x-1}+\frac{C}{\left(x-1\right)}+\frac{D}{{\left(x-1\right)}^2}$

$\Rightarrow \frac{\left(5x^2-4x+1\right)}{{\left(x-1\right)}^2}=B+\left(2x-1\right)\left\{\frac{C}{\left(x-1\right)}+\frac{D}{{\left(x-1\right)}^2}\right\}$  Putting $x=\frac{1}{2}$ we get $B=1$.

Again,

$\frac{\left(5x^2-4x+1\right)}{\left(2x-1\right){\left(x-1\right)}^2}=\frac{B}{2x-1}+\frac{C}{\left(x-1\right)}+\frac{D}{{\left(x-1\right)}^2}$

$\Rightarrow \frac{\left(5x^2-4x+1\right)}{\left(2x-1\right)}=\left\{{\left(x-1\right)}^2\times \frac{B}{2x-1}\right\}+\left(x-1\right)C+D$

Putting $x=1$, we get $D=2$.

Now, considering the actual expression, we have

$\frac{x^3}{(2x-1)(x-1{)}^2}=A+\frac{B}{2x-1}+\frac{C}{\left(x-1\right)}+\frac{D}{{\left(x-1\right)}^2}$

Putting $x=0$, we get

$A-B-C+D=0$

$\Rightarrow \frac{1}{2}-1-C+2=0$

$\Rightarrow C=\frac{3}{2}$

Now,

$2A-3B+4C+5D$

$\Rightarrow 2\left(\frac{1}{2}\right)-3\times \left(1\right)+4\left(\frac{3}{2}\right)+5\times \left(2\right)=14$