Where is the error in this partial fraction decomposition?

Question

$$\frac{1}{(2)(u)(\sqrt{u}-1)}$$ $$\frac{1}{(2)(u)(\sqrt{u}-1)} = \frac{A}{2} + \frac{B}{u} + \frac{C}{\sqrt{u}-1}$$ $$1 = Au\sqrt{u} - Au + 2B\sqrt{u}-2B+2Cu$$ $$1 = u\sqrt{u}(A)+u(-A-2C)+\sqrt{u}(B)-2B$$ $-2B = 1\Rightarrow B=-\frac{1}{2}$ since this is the only constant value on the right side and so must equal one. $A = 0$ since there is no term on the left side with u with the exponent $\frac{3}{2}$ ($u\sqrt{u} = u^\frac{3}{2}$). Thus, $-A-2C=0$ (no term with u in it) $\Rightarrow 0-2C=0 \Rightarrow C=0$. Now we get: $$\frac{1}{(2)(u)(\sqrt{u}-1)} = \frac{0}{2} + \frac{-\frac{1}{2}}{u} + \frac{0}{\sqrt{u}-1}$$ $$\frac{1}{(2)(u)(\sqrt{u}-1)} = -\frac{1}{2u}$$ What went wrong?

Answer 1

You have to have ${B\sqrt{u}+C\over u}$ instead of ${B\over u}$ in second line. You would notice this quicker if you wrote $x=\sqrt{u}$.