Performing A Partial Fraction Decomposition

Question

I have the following fraction and I want to decompose it in partial fractions:

$$\dfrac{1}{(x^2-1)^2}$$

Here is how I started the decomposition:

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

• $A = \dfrac{(x-1)^2}{(x-1)^2(x+1)^2} - B(x-1) - \dfrac{C(x-1)}{(x+1)^2}-\dfrac{D(x-1)^2}{x+1}\biggr\rvert_{x=1} = \dfrac{1}{4}$

• Similarly we get $C = \dfrac{1}{4}$

Now my problem is I don't know how to find $B$ and $D$. If I had only 1 unknown value It would have been fine but there are 2.

Answer 1

That values are correct. But, to find the else two you have to rewrite this as:

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

Then multiply all this and let the values near $x^3, x^2, x$ be $0$, and the free part to be $1$. Using that and values of $A, C$ you can easy find`$ B, D$.

You will get $B= -\frac{1}{4}$ and $D = \frac{1}{4}$.

Answer 2

First, as the function is even, you can easily check $C=A$, $D=-B$.

Second, transform your equality into a polynomial quality, it will be easier to work with: taking into account the above observation, we get, multiplying both sides with $(x^2-1)^2$: $$1=A(x+1)^2+B(x-1)(x+1)^2+A(x-1)^2-B(x-1)^2(x+1). $$ Setting $x=0$ yields $\;1=2A-2B$, whence $\;B=-\dfrac14$.