Partial fraction decomposition of even function

Question

I need to do partial fraction decomposition of this function (to solve its integral):

$\frac{t^2}{t^4+4}$

Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:

$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$

And then continue with calculations.

According to the teacher's notes though, I can instead write it like this because the function is even:

$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}-A\frac{2t-2}{t^2-2t+2}+B\frac{1}{t^2-2t+2}$

But I can't understand how to get it, knowing that the function is even.

Can you please help me?

Answer 1

Since the function is even we have that

$$A\frac{2t+2}{(t+1)^2}+B\frac{1}{(t+1)^2}+C\frac{2t-2}{(t-1)^2}+D\frac{1}{(t-1)^2}$$ is equal to

$$A\frac{-2t+2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}.$$ Reording terms we get

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

Comparing the first and third expressions we get $A=-C, B=D.$

Answer 2

Let $t=2u$

$$\dfrac{2u^2}{u^4+1}=\dfrac{u^2-1+u^2+1}{u^4+1}=\dfrac{1-1/u^2}{(u+1/u)^2-2}+\dfrac{1+/u^2}{(u-1/u)^2+2}=?$$