Expressing $\frac{1}{4n^2-1}$ as a partial fraction

Question

I was asked to express $$\frac{1}{4n^2-1}$$ as a partial fraction. I have no clue as to what I should break this into. For example I know :

$$\frac{1}{n(n-1)}= \frac {A}{n} + \frac {B}{n-1}$$

These are fractorisable. Whereas $4n-1$ isnt.

So how to do it when one can't factorize the denominator?

Answer 1

$$\frac{1}{4n^2-1}=\frac{1}{(2n)^2-1}=\frac{1}{(2n+1)(2n-1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$$

$A=0.5$, $B=-0.5$

Answer 2

$$\frac{1}{4n^2-1}=\frac{1}{(2n)^2-1}=\frac{1}{(2n+1)(2n-1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$$

$$1=A(2n+1)+B(2n-1)$$ Set $n=-.5$$$1=A(0)+B(-2)$$ $$B=-\frac12$$ Set $n=.5$ $$1=A(2)+B(0)$$ $$A=\frac12$$

$$=\frac1{4n-2} - \frac1{4n+2}\quad \blacksquare$$

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If you are doing this for integration:

$$\int \frac{.5}{2n-1} dn-\int\frac{.5}{2n+1} dn$$ $$\frac14\ln|2n-1| - \frac14\ln|2n+1|+C$$ $$\frac14\left(\ln\left|\frac{2n-1}{2n+1}\right|\right)+C\quad\blacksquare$$

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If you are doing limits:

$\lim \limits_{n\to\pm\infty} \frac1{4n^2-1} = 0$

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If you are doing series: