Dealing with degrees in decomposition into partial fractions I had to decompose $ \frac{2x^2}{x^4-1} $ into partial fractions in order to determine its antiderivative.
So, I said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}
$$
However, in the answer key, they said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x^2-1} + \frac{B}{x^2+1}
$$
Although I got the same answer, I had to do unpleasant calculations to finally get a system of four equations and four unknowns, which I had to solve as well.
What I want to know is what made them do that assumption? This isn't what we learned about decomposition into partial fractions.
I thought that maybe because in the starting fraction, the polynomial in the numerator is 2 degrees less than that of the denominator, so we must keep the same degree difference in the partial fractions.
However if we look at this example where the numerator is 5 degrees less than the denominator, this what was written in the answer key:
$$
\frac{1}{x^2(x-1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}
$$
Which is normal and compatible with what I've learned about decomposition into partial fractions.
Please can anyone help? Also please no very complex calculations because I am a biology student. Thank you.

In your very first example, notice that the function
$$f(x)=\frac{2x^2}{x^4-1}$$
is an even function (a function that satisfies $f(-x)=f(x)$ for all $x$). So, the partial fraction descomposition 
$$g(x)=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}$$
also satisfies the same. Thus
$$\begin{align} 
\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}
&= \frac{a}{-x-1}+\frac{b}{-x+1}+\frac{-cx+d}{x^2+1} \\
&=-\frac{a}{x+1}-\frac{b}{x-1}+\frac{-cx+d}{x^2+1}
\end{align}$$
Can you see this implies that $c=0$ and we can consider only a one variable in the top of $(x+1)(x-1)=x^2-1$?
However, in the second example, the function is not even neither odd, so, you have to do the decomposition in the traditional way.
