Partial fraction decomposition of even function 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?

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 $A=-C, B=D.$
