# 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?

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$$\frac{1}{1+x} \frac{1}{1-x}\neq \frac{1-x}{1-x}\frac{1}{1+x} - \frac{1+x}{1+x}\frac{1}{1-x}$$ (Try $x=0$). –  Git Gud Apr 3 '13 at 22:22

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 $1-x=-(x-1)$.

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D'oh, of course. The example that I'm working from does have the errant formula, but the intent should have been obvious to me. Thanks! –  Matthew Apr 3 '13 at 22:25
@Matthew It is not your fault! –  Pedro Tamaroff Apr 3 '13 at 22:28

you wrote wrong fraction : $$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1-x}-\dfrac{1}{1+x})$$

instead of :$$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1+x}-\dfrac{1}{1-x})$$ Now check it by plug in $x=2$

LHS=$\dfrac {1}{-3}\,$ ; RHS=$\dfrac{1}{2}(\dfrac{1}{3}-{1})\implies \dfrac {1}{-3}$

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