# Intuition behind partial fractions

I was trying to solve the following integral:

$$\int \frac{dx}{(x^2 + 2x + 1)(x^2 + 1)}$$

But I'm having some trouble doing partial fractions decomposition with this one. What I did was the following: Firstly I simplified the denominator of the fraction:

$$\int \frac{dx}{(x^2 + 1)(x + 1)^2}$$

So, I know that, to do partial fractions with this we are suposed to write it as:

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{*}{x^2 + 1} + \frac{*}{x + 1} + \frac{*}{(x + 1)^2}$$

The part I'm having touble with is figuring out what should go in the $$*$$. I checked in WolframAlpha and it's supossed to be written as:

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 1} + \frac{D}{(x + 1)^2}$$

My question is: How can I conclude this? Is there a general rule for doing this? What is the thought process behind choosing what to put in the numerator of the fractions? My teacher only showed us some examples and I also find myself having some trouble doing this when it's a type of fraction I've never seen before, can you give me some general tips when doing this?

• In general, the partial fraction $P/Q$ has the following property: the degree of $P$ must be one less than the degree of $Q$. So we have a constant over a linear, a linear over a quadratic, a quadratic over a cubic, etc. If the degree of $P$ is strictly less than $Q$, then it is a proper fraction. Otherwise, it is an improper fraction, and we need to use polynomial long division.
– Joe
Jan 5, 2021 at 16:09
• In this case would $(x+1)^2$ count as a quadratic or a linear? Because WolframAlpha did: $C/(x+1)^2$ @Joe Jan 5, 2021 at 16:13
• It is in a sense a linear expression—just with the coefficient of $x$ being $0$.
– Joe
Jan 5, 2021 at 16:15
• The polynomials $x^2+1$ and $x+1$ are irreducible, they cannot be factored over the reals. The general rule is that if the denominator is irreducible or a power of an irreducible, take the numerator to be a polynomial of degree one less than the degree of the irreducible. The reason for the rule lies in the theory of partial fractions, which reduces the original fraction to the sum of fractions of the form $p/q^n,$ with $q$ an irreducible. As Joe mentions, if the degree of $p$ is greater than that of $q$, you can do polynomial long division to reduce to the case in which ... Jan 5, 2021 at 16:37
• The reason wolfram wrote only a constant in the last slot because if it were linear then $$\frac{Cx+D}{(x+1)^2} = \frac{Cx+C-C+D}{(x+1)^2} = \frac{C}{x+1}+\frac{D-C}{(x+1)^2}$$ which, since the constants are arbitrary, makes it equivalent to the above formula. Jan 5, 2021 at 16:41

Rule of thumb: "The partial fraction $$\frac{P(x)}{Q(x)}$$ has the following property: the degree of $$P(x)$$ must be one less than the degree of $$Q(x)$$. So we have a constant over a linear, a linear over a quadratic, a quadratic over a cubic, etc." and this always works (Taken from Joe's comment)

Actually, you take linear in the numerator even for $$(x+1)^2$$ but it gets reduced

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C}{x + 1} + \frac{Dx+E}{(x + 1)^2}$$

$$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C}{x + 1} + \frac{D(x+1)+(E-D)}{(x + 1)^2}$$ $$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{Ax+B}{x^2 + 1} + \frac{C+D}{x + 1} + \frac{(E-D)}{(x + 1)^2}$$

And you get the form, $$\frac{1}{(x^2 + 1)(x + 1)^2} = \frac{*}{x^2 + 1} + \frac{*}{x + 1} + \frac{*}{(x + 1)^2}$$

So the rule of thumb always works, WolframAlpha gives you a simplified version.

Hope this helped.:)

• Maybe you mean to leave out $C/(x+1)$? (I voted +1 BTW) Jan 5, 2021 at 16:45

If I were you I would get an online reference for partial fractions. It isn't hit or miss.

The theory works out pretty much goes as follows.

1. Assuming $$p$$ and $$q$$ polynomial, and $$\deg{p}<\deg{q}$$, factorise $$q$$ fully. Over the reals you can factor into a product of linear and quadratic terms. Ensure that you group terms in the sense that $$(x+1)(2x+2)=2(x+1)^2$$ or $$(x^2+1)(x^2+1)(4x^2+4)=4(x^2+1)^3$$, etc. (basically group terms with same roots).

2. Now, each factor in the factorisation begets a term in the partial fraction expansion according to four rules.

I: a single linear factor $$ax+b$$ begets a $$A/(ax+b)$$. II: a repeated linear factor $$(ax+b)^n$$ begets:

$$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_n}{(ax+b)^n}.$$

Note $$(ax+b)^n$$ is $$n$$ factors multiplied together and so there are $$n$$ terms.

This reduces to rule I if $$n=1$$.

III: a quadratic $$ax^2+bx+c$$ begets a term $$(Ax+B)/(ax^2+bx+c)$$

and rule IV follows a similar pattern to Rule II.

WHY this works I would suggest you find a reference for that. The one thing that might help you understand is the following:

Say your $$1/((x^2+1)(x+1)^2)$$... so you have a $$x^2+1$$ and a $$x+1$$ and another $$x+1$$. What won't work is:

$$\frac{1}{(x^2+1)(x+1)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x+1}$$

because those last two terms give $$(C+D)/(x+1)$$... and there are examples where you cannot get the rational function on the left here without a $$1/(x+1)^2$$ term... it just doesn't work. So you change the $$D/(x+1)$$ to $$D/(x+1)^2$$ and the theory (to be found in a reference) says that this will always work.

Hopefully this is some kind of help. It isn't a scatter gun approach, there is theory behind the rules.