How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions

Question

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

Answer 1

The form you chose $$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$ is not general enough - your calculation proves this. It is clear apriori, since why should $1$ be expressible in the form $A(1-x)^2 + B(1+x)$? There are $3$ constraints (the coefficient of $x^2$ should be $0$, as well as that of $x$, and the constant term should be $1$), but you've allowed only $2$ degrees of freedom $A,B$.

The general form which allows decomposing any fraction into partial fractions can be found e.g. in Wikipedia (see specifically the examples where it is the easiest to understand it).

Specifically for your problem, the form you should try to decompose into is

$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{1 - x} + \frac{C}{(1-x)^2}.$$

Answer 2

try this one $$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{Bx+C}{(1 - x)^2} $$ and find $A,B,C$

Answer 3

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, $$\frac{1}{(1+x)(1-x)^{2}}=\frac{-1/4}{1-x}+\frac{1/2}{(1-x)^{2}} +\frac{1/4}{1+x}.$$

Answer 4

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

$$\frac{1}{(1 + x)(1 - x)^2}=\frac 12 \frac{1}{(1 - x)^2} + \frac 14 \frac{1}{1 - x} + \frac 14 \frac{1}{1 + x}$$

Answer 5

Hint: $$1=\frac{(1+x)+(1-x)}2.$$ This should make the exercise trivial.