# 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.

• This is a way of doing partial fractions that is more like looking for symmetries than grinding out systems of equations. Instead of taking the factoring all the way, one could take it one step to $x^4-1 = (x^2-1)(x^2+1)$ and realize that both terms contain the $x^2$ in the numerator of the original, so is there any combination of the two that adds to the original? With a quick check, you see that $\frac{1}{x^2-1}+\frac{1}{x^2+1} = \frac{x^2+1+x^2-1}{(x^2-1)(x^2+1)} = \frac{2x^2}{x^4-1}$. Commented Oct 19, 2019 at 18:30

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.

• Aha yes I see! By comparison, right? I can't upvote your answer since i don't have enough reputation, but thank you! Commented Oct 19, 2019 at 18:47

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

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 $$\int\frac{\mathrm dx}{1-x^2}=\tfrac12\ln\biggl|\frac{1+x}{1-x}\biggr|=\operatorname{argtanh}x.$$

• Okay, thank you! And yes I know inverse hyperbolic functions. Commented Oct 19, 2019 at 18:49