Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$. 
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.

Setting $x=\sqrt 2i$ (so $x^2=-2$) yields
$$-4D-E+\sqrt2(-2D+E)i=2+\sqrt 2 i,\quad\text{whence }\begin{cases}\;2D+E=-2,\\-D+E=1,
\end{cases}\iff D=-1,\;E=0.$$
Next, setting $x=0$, you get
$$2=4A-2C-E, \quad\text{ so }\; C=-\tfrac13.$$
Last, multiply both sides of the decomposition by $x$ and let $x\to+\infty$, obtaining:
$$1=A+B,\quad\text{ so }\;B=\tfrac23.$$
