<s>[INST] partial fractions 
$$\frac{1}{1-x^2}$$

$$\frac{1}{1-x^2}=\frac{a}{1-x}+\frac{b}{1+x}$$
$$1=a+ax+b-bx$$
$$1=a+b+x(a-b)$$
$a+b=1$ and $x(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$
$$2a=1\Rightarrow a=\frac{1}{2}$$
$b=\frac{1}{2}$
$$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
Where I went wrong?
 [/INST] You didn't do anything wrong.
$$\frac{1}{1-x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)}$$
is equivalent to your answer of
$$\frac{1}{1-x^2} = \frac{1}{2(1-x)} + \frac{1}{2(1+x)}.$$
I think we can agree that both answers have a common term of $\dfrac{1}{2(1+x)}$.  Now, notice:
$$\frac{1}{2(1-x)} = \frac{1}{2 \cdot [-1(x-1)]} = \frac{1}{-2(x-1)} = -\frac{1}{2(x-1)}$$
Therefore $-\dfrac{1}{2(x-1)} = \dfrac{1}{2(1-x)}$, and so the answers are the same.
</s>