What's the best way to solve $1 = A(x^2+1) + (Bx+C)(x+1)$ What's the best way to solve $$1 = A(x^2+1) + (Bx+C)(x+1)$$
I let $x=-1$ and got $A=\frac{1}{2}$
But what sub is ideal to find B&C
This gets messy quick, I think.
Instead, I started over, and I grouped like terms 
$$1 = Ax^2+A+Bx^2+Bx+Cx+C$$
This led to $$A+B=0$$ $$B+C=0$$ $$A+C=1$$
This led to $B=-\frac{1}{2}$ 
$C=\frac{1}{2}$
and solved using my A=1/2
But, I did not like that I had to mix techniques.

At each step take advantage of your previous findings.$$1 = A(x^2+1) + (Bx+C)(x+1)$$
$$x=-1 \implies A=\frac {1}{2}$$
$$ x=0 \implies A+C=1 $$
Thus $$C =1/2$$
$$x=1 \implies 2A+2(B+C)=1$$ 
Thus $$B+C=0 \implies B=-\frac {1}{2}$$ 
