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?
 A: 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.
A: 
Notice:

*

*$$\frac{1}{1-x^2}=\frac{1}{-(x-1)(x+1)}=-\frac{1}{(x-1)(x+1)}$$

So, we get:
$$-\frac{1}{(x-1)(x+1)}=-\frac{a}{x-1}-\frac{b}{x+1}$$
And, now we can see that:
$$a(x+1)+b(x-1)=1\Longleftrightarrow a-b+(a+b)x=1$$
So, we get:

*

*$$a-b=1\Longleftrightarrow a=b+1\Longleftrightarrow a=-\frac{1}{2}+1=\frac{1}{2}$$

*$$a+b=0\Longleftrightarrow b+1+b=0\Longleftrightarrow 2b=-1\Longleftrightarrow b=-\frac{1}{2}$$
So, now we have the partial fractions:
$$\frac{1}{1-x^2}=-\frac{\frac{1}{2}}{x-1}-\frac{-\frac{1}{2}}{x+1}=\frac{1}{2(x+1)}-\frac{1}{2(x-1)}$$
A: But you are not wrong.
$$\frac{1}{2(1-x)}+\frac{1}{2(1+x)} = \frac{1+x+1-x}{2(1+x)(1-x)} = \frac{2}{2(1+x)(1-x)} = \frac{1}{1-x^2}$$
A: Whats the problem take $1/2$ common so you get $1/2(\frac{1+x+1-x}{1-x^2})=\frac{2}{2(1-x^2)}=\frac{1}{1-x^2}$
