Intuition behind partial fractions

Question

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?

Answer 1

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.:)

Answer 2

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.