Dealing with degrees in decomposition into partial fractions

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

They simply used that the given function is even, so that, since $f(x)=f(-x)$ for all $x$, one deduces, with your notations, $$B=-A, \qquad C=0$$ and there really remains only two unkown coefficients.

Furthermore, you don't have to decompose $\;\dfrac A{x^2-1}$, since you're supposed to know by heart (well… maybe not for a biology student) that $$\int\frac{\mathrm dx}{1-x^2}=\tfrac12\ln\biggl|\frac{1+x}{1-x}\biggr|=\operatorname{argtanh}x.$$