partial fractions when the fraction cannot be decomposed

Question

I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can I then go about writing this fraction as the sum $\frac {1}{2}\left [\frac {1}{x^2+1}-\frac{x^2-1}{(x^2+1)^2}\right]$? Is it just manipulation and trial and error?

Since $i^2 = -1$, you could write the denominator as $(x^2 - i^2) = (x+i)(x-i)$. — Khallil, Dec 21 '20 at 12:00
Why do you want to find the partial fractions? If you want to integrate this, just substitute $x = \tan u, dx = \sec u \ du$. — Toby Mak, Dec 21 '20 at 12:18
@TobyMak: there's more to (mathematical) life than to blindly find antiderivatives. — Martin Argerami, Dec 21 '20 at 13:03
That's true, but it's still a possibility. I worry this may be one of those XY problems where the OP gets an answer but the true issue lies unresolved. — Toby Mak, Dec 21 '20 at 13:05
See also https://math.stackexchange.com/questions/35924/integral-of-frac11x22. — Hans Lundmark, Mar 17 '21 at 18:28

score 3 · Answer 1 · edited Dec 21 '20 at 12:24

3

It can be decomposed if you use the complex roots $$(x^2+1)^2=(x-i)(x+i)(x-i)(x+i)$$ $$\frac 1{(x^2+1)^2}=\frac{i}{4 (x+i)}-\frac{1}{4 (x+i)^2}-\frac{i}{4 (x-i)}-\frac{1}{4 (x-i)^2}$$

edited Dec 21 '20 at 12:24

Adam Rubinson

20,052

answered Dec 21 '20 at 12:01

Claude Leibovici

260,315

score 3 · Accepted Answer · answered Dec 21 '20 at 12:19

$\frac{1}{(x^2+1)^2}$ is already in "partial fractions form", so...

"Is it just manipulation and trial and error?"

Yes.

It's not super trivial, but the result is true.

I did:

\begin{align} \frac{1}{(x^2+1)^2} &= \frac{1 + x^2 - x^2}{(x^2+1)^2} \\\\ &= \frac{1}{(x^2+1)}\ - \frac{x^2}{(x^2+1)^2} \\\\ &= \frac12\left[\frac{2}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right] \\\\ &= \frac12\left[\frac{1}{(x^2+1)} + \frac{1}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right]\\\\ &= \frac12\left[\frac{1}{(x^2+1)}\ + \frac{x^2+1-2x^2}{(x^2+1)^2}\right] \\\\ &= \frac12\left[\frac{1}{(x^2+1)}\ - \frac{x^2-1}{(x^2+1)^2}\right]. \\\\ \end{align}

GEdgar · Answer 3 · 2021-06-25T18:26:22.100

In the real-valued partial fraction method, your term $$ \frac{1}{(x^2+1)^2} $$ is the best you can do. Evaluation of integrals $$ \int\frac{dx}{(ax^2+bx+c)^n} $$ is done as follows: For $n=1$, complete the square and it is an arctangent. For $n>1$ there is a reduction formula from integration by parts: \begin{align} \int\frac{dx}{(ax^2+bx+c)^n} &= \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}} \\ &\qquad+\frac{(2n-3)2a}{(n-1)(4ac-b^2)} \int\frac{dx}{(ax^2+bx+c)^{n-1}} \end{align}

score 2 · Answer 4 · answered Dec 21 '20 at 16:34

Just to clarify:

In cases of irreducible denomiantors over the real numbers such as your case, we treat it like a regular fraction with a repeated factor in the denominator but instead of letting the numerators of each fraction produced by the partial fraction decomposition be constant, instead we let the numerator be a polynomial of degree one lower than the factor in question.

That was rather a mouthful! To clarify, in your case:

$$\frac{1}{(x^2+1)^2}\equiv\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$$

Finding those coefficients, however, yields $D=1$ and this is the only non zero coefficient. This is why your fraction is already in partial fraction form.

partial fractions when the fraction cannot be decomposed

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