How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks

A direct splitting as follows is also possible and sometimes quite quick:
\begin{eqnarray*} \color{blue}{\frac{1}{(1 + x)(1 - x)^2}}
& = & \frac{1+x-x}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1-x -1}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1}{(1 + x)(1 - x)} \color{blue}{- \frac{1}{(1 + x)(1 - x)^2}} \\
\end{eqnarray*}
Hence,
\begin{eqnarray*} \frac{\color{blue}{2}}{(1 + x)(1 - x)^2}
& = & \frac{1}{(1 - x)^2} + \frac 12\frac{1+x+1-x}{(1 + x)(1 - x)}\\
\end{eqnarray*}
So, you get
$$\frac{1}{(1 + x)(1 - x)^2}=\frac 12 \frac{1}{(1 - x)^2} + \frac 14 \frac{1}{1 - x} + \frac 14 \frac{1}{1 + x}$$
