Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.

Question

Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.

My method:

$\frac{9x}{(2x+1)^2(1-x)}\equiv\frac{A}{2x+1}+\frac{B}{(2x+1)^2}+\frac{C}{1-x}$

Comparing L.H.S. numerators with R.H.S. numerators: $9x=A(2x+1)+B(1-x)+C(2x+1)^2$

$x=-\frac12, B=-3$

$x=0, A+B+C=0 \Rightarrow A=3+C$

$x=1, 3A+9C=9 \Rightarrow 3(3+C)+9C=9\Rightarrow C=0$

This is the point where my answer differs from the one given in the book. And I don't understand why.

Answer in book: $\frac{1}{1-x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}$

Answer 1

You have to have $$9x=A(2x+1)\color{red}{(1-x)}+B(1-x)+C(2x+1)^2.$$

Answer 2

Spotted mistake. $A(2x+1)$ should be $A(2x+1)(1-x)$