Separating $\frac{1}{1-x^2}$ into multiple terms I'm working through an example that contains the following steps:
$$\int\frac{1}{1-x^2}dx$$
$$=\frac{1}{2}\int\frac{1}{1+x} - \frac{1}{1-x}dx$$
$$\ldots$$
$$=\frac{1}{2}\ln{\frac{1+x}{1-x}}$$
I don't understand why the separation works. If I attempt to re-combine the terms, I get this:
$$\frac{1}{1+x} \frac{1}{1-x}$$
$$=\frac{1-x}{1-x}\frac{1}{1+x} - \frac{1+x}{1+x}\frac{1}{1-x}$$
$$=\frac{1-x - (1+x)}{1-x^2}$$
$$=\frac{-2x}{1-x^2} \ne \frac{2}{1-x^2}$$
Or just try an example, and plug in $x = 2$:
$$2\frac{1}{1-2^2} = \frac{-2}{3}$$
$$\frac{1}{1+2} -\frac{1}{1-2} = \frac{1}{3} + 1 = \frac{4}{3} \ne \frac{-2}{3}$$
Why can $\frac{1}{1-x^2}$ be split up in this integral, when the new terms do not equal the old term?

The thing is $$\frac{1}{1-x}\color{red}{+}\frac 1 {1+x}=\frac{2}{1-x^2}$$
What you might have seen is 
$$\frac{1}{x-1}\color{red}{-}\frac 1 {x+1}=\frac{2}{1-x^2}$$
Note the denominator is reversed in the sense $1-x=-(x-1)$.
