Partial fractions - different results when done in steps than not We have: $\frac 1 {(1-x)(1+x)(1-2x)}$
If I do the partial fractions straight: $\frac 1 {(1-x)(1+x)(1-2x)}= \frac a {1-x} + \frac b {1+x} + \frac c {1-2x}$
I get: $a=-\frac 12, b = \frac 1 6, c=\frac 4 3$.
But when I do it in steps, i.e: $\frac 1 {(1-x^2)(1-2x)}=\frac a {1-x^2} +  \frac b {1-2x} $
I get: $a=2, b=-1$, then again partial fractions: $\frac 2 {1-x^2}=\frac a {1-x} + \frac b {1+x}$ Get: $a=1, b=-1$
In the first method I got: $\frac {-\frac 12} {1-x} + \frac {\frac 1 6} {1+x} + \frac {\frac 4 3} {1-2x}$
But in steps: $\frac {1} {1-x} - \frac {1} {1+x} - \frac {1} {1-2x}$
How is that possible? Shouldn't they be the same?

$\frac 1 {(1-x^2)(1-2x)}\ne\frac a {1-x^2} +  \frac b {1-2x}$
$\frac 1 {(1-x^2)(1-2x)}=\frac {a+bx} {1-x^2} +  \frac c {1-2x}$
One finds that $a=-1/3$, $b=-2/3$ and $c=4/3$.
