<s>[INST] 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
 [/INST] Using the hint given by @GEdgar, notice that
$$\frac{1}{(1+x)(1-x)^{2}}=\frac{A}{1-x}+\frac{B}{(1-x)^{2}} +\frac{C}{1+x}.$$
Solving, we get
$$1=A(x^{2}-1)+B(x+1)+C(x-1)^{2}$$
Equality coefficients, we have
$$\begin{cases}-A+B+C=1,\\  0A+B-2C=0,\\A+0B+C=0\end{cases}$$
Solving the system equations, we get
$$\left(A,B,C\right)=\left(-\frac{1}{4},\frac{1}{2}, \frac{1}{4}\right).$$
Therefore,
$$\frac{1}{(1+x)(1-x)^{2}}=\frac{-1/4}{1-x}+\frac{1/2}{(1-x)^{2}} +\frac{1/4}{1+x}.$$
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