<s>[INST] 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.
 [/INST] There are always alternatives, one does not expect an answer key to give every way that works. For finding an antiderivative, 
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}
$$
is quite good in terms of recognizing things. $C$ leads to a substitution, while $D$ leads to an arctangent
</s>