Partial Fraction Decomposition (Linear Denominators)
Key Questions

Answer:
Transforming a rational polynomial into a sum of simpler rational polynomials due to the factorization of the denominator.
Explanation:
Let me try and explain this to you. :)
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What is a partialfraction decomposition?
We are all familiar with the process of adding and subtracting fractions, e.g.
#1/(x+1) + 2/(x+3) = (x+3)/((x+1)(x+3)) + (2(x+1))/((x+1)(x+3)) = (3x + 5)/(x^2 + 4x + 3)# Now, partialfraction decomposition does exactly the reverse thing.
It takes a rational polynomial, so a fraction like
#(3x + 5)/(x^2 + 4x + 3)# from my example, and tries to decompose it into a sum of "simpler" fractions (to be more precise: into a sum of fractions which denominators are factors of the original fractions's denominator.)
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How to compute a partialfraction decomposition?1) Linear and unique factors
Let's stick with my example:
#(3x + 5)/(x^2 + 4x + 3)# The first thing to do is a always to find a complete factorization of the denominator:
#x^2 + 4x + 3 = (x + 3)(x + 1)# Here, all the factors are linear and unique, this is the simple case.
The goal is to to find
#A# ,#B# so that the following equation holds:#(3x + 5)/((x+3)(x+1)) = A/(x+3) + B / (x+1)# To do so, first we should multiply both sides with the denominator:
#<=> 3x+ 5 = A ( x+1) + B ( x+3)# Now, in order to solve this equation, we need to "group" the
#color(red)(x)# terms and the terms#color(blue)("without "x)# (more precisely:#x^0# terms):#color(red)(3x)+ color(blue)(5) = color(red)(A * x) + color(blue)(A) + color(red)(B * x) + color(blue)(3 B)# Thus, we can form an equation based on the "red" terms and an equation based on the "blue" terms:
#{ (3 = A + B), (5 = A + 3B):}# The solution of this linear equation system is inded
#A = 2# and#B = 1# which leads us to the following partialfraction decomposition:#(3x + 5)/((x+3)(x+1)) = A/(x+3) + B / (x+1) = 2/(x+3) + 1/(x+1)# ++++++++++++++++++++++
2) Nonunique factors
The computation is more complicated in case that the complete factorization of the original denominator doesn't consist just of linear and unique factors.
Just a short example:
Let's say we could factorize our rational polynomial like follows:
#x/(x^2(x+1)^2(x1)# As the factors
#x# and#x+1# occur twice in the denominator, we need to account for that by building fractions with increasing powers of those factors.
In this case, we would need to find#A# ,#B# ,#C# ,#D# and#E# so that#x/(x^2(x+1)^2(x1)) = A/x + B/x^2 + C/(x+1) + D/(x+1)^2 + E/(x1)# holds.
++++++++++++++++++++++
3) Nonlinear factors
Last but not least, there might be nonlinear factors in your factorized denominator.
Example:
#1/(x^3+x) = 1/(x(x^2+1))# Unfortunately, you can't factorize
#x^2+1# further (at least, for real numbers).In this case, your partialfraction decomposition needs to look like follows:
Find#A# ,#B# ,#C# so that the following equation holds:#1/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
When can a partialfraction decomposition be useful?Finally, I'd like to note that a very common application of this method is integration of rational polynomials:
If you need to compute
#int (3x + 5)/(x^2 + 4x + 3) "d"x# and know that
#1/(x+1) + 2/(x+3) = (3x + 5)/(x^2 + 4x + 3)# holds,it's easy to integrate by "splitting" the original fraction into a sum of simpler fractions and integrating each one of them:
#int (3x + 5)/(x^2 + 4x + 3) "d"x = int 1/(x+1) "d"x + int 2/(x+3) "d"x#