Step 1: Clear the denominators.
Multiply both sides by $(x^2+1)^2(x-1)$ to get \[ x=(Ax+B)(x^2+1)(x-1)+(Cx+D)(x-1)+E(x^2+1)^2. \]
Step 2: Find $E$ quickly.
Put $x=1$. The first two terms vanish, and $(x^2+1)^2=4$, so $1=4E$, giving $E=\dfrac14.$
Step 3: Compare highest power.
The $x^4$ terms come from $A x\cdot x^2\cdot x=Ax^4$ and $E x^4$. Their sum must be $0$ (no $x^4$ on the left): $A+E=0$, so $A=-\dfrac14.$
Step 4: Get $B$, $C$, $D$ by matching more powers.
Comparing the $x^3,x^2,x^0$ coefficients in turn and using $A=-\tfrac14,E=\tfrac14$, we solve the small system to find $B=-\dfrac14,\ C=-\dfrac12,\ D=\dfrac12.$
Step 5: Plug into the asked expression.
\[ A+B-C+2D=-\frac14+\left(-\frac14\right)-\left(-\frac12\right)+2\left(\frac12\right). \]
Step 6: Add carefully.
$=-\dfrac14-\dfrac14+\dfrac12+1=-\dfrac12+\dfrac12+1=1.$ \[ \boxed{1} \]