Fraction decomposition

Integration: Integration techniques

Fraction decomposition

We have seen how we can decompose fractions with several terms in the numerator into a sum of several fractions with the same denominator. We noted that it is not always allowed to decompose the denominator. Yet there are situations where this is possible. This can, among other things, help to integrate this type of fraction.

Below are some examples of situations in which fractions can be decomposed.

\[\begin{array}{rcl}\dfrac{7x+31}{(x+3) \cdot (x+5)}&=&\dfrac{5}{x+3}+\dfrac{2}{x+5} \\ \\ \dfrac{6x-4}{x\cdot (x-2)}&=&\dfrac{2}{x}+\dfrac{4}{x-2} \\ \\ \dfrac{12x+15}{x^2+x-2}&=& \dfrac{9}{x-1}+\dfrac{3}{x+2}\end{array}\]

When we look closely at these examples, we see that if we multiply the denominators on the right-hand side, we get the denominator on the left-hand side.

We will now look at how we can decompose a fration of the form #\frac{\purple l \cdot x+\orange m}{x^2+bx+c}#, in which the the denominator #x^2+bx+c# can be written as #(x+\blue p) \cdot (x+\green q)#.

Step-by-step fraction decomposition

	Step-by-step	Example
	We decompose a fraction of the form #\frac{\purple l \cdot x+\orange m}{x^2+bx+c}# in which #x^2+bx+c# can be written as #(x+\blue p)\cdot (x+\green q)#.	#\frac{1}{x^2+5x+6}#
Step 1	Write #x^2+bx+c=(x+\blue p)\cdot (x+\green q)#.	#x^2+5x+6=(x+\blue 2) \cdot (x+\green 3)#
Step 2	Write #\frac{\purple l \cdot x+\orange m}{x^2+bx+c}=\frac{A}{x+\blue p}+\frac{B}{x+\green q}#	#\frac{\orange1}{x^2+5x+6}=\frac{A}{x+\blue 2}+\frac{B}{x+\green 3}#
Step 3	Bring the right-hand side together in one denominator and compare the numerators on both sides. This gives the equation: \[\purple l \cdot x+\orange m=A\cdot(x+\green q)+B \cdot (x+\blue p)\]	\[\orange1=A \cdot (x+\green3) + B \cdot (x+\blue2)\]
Step 4	Simplify the equation from step 3 by expanding the brackets, and combining the terms with #x#.	\[\orange1=(A+B) \cdot x+\green3A+\blue2B\]
Step 5	Compose a system of linear equations by comparing the coefficients of the left and right terms of the equation in step 4.	\[\lineqs{A+B&=&\purple0\cr \green3A+\blue2B&=&\orange1\cr}\]
Step 6	Solve the system from Step 5 using substitution or elimination.	\[\lineqs{A=1 \cr B=-1}\]
Step 7	Substitute the values found in step 6 in the right-hand side of the equation of step 2 in for the answer.	#\frac{1}{x+\blue2}+\frac{-1}{x+\green3}#

p=q

When the quadratic denominator can be written as #(x+\blue p)^2#, it is also possible to decompose the fractions. This works the same way, only we must apply a little trick by splitting step 2 into #\frac{A}{x+\blue p} + \frac{B}{(x+\blue p)^2}#. We can then use the steps below, in which the difference can be found in step 2.

	Step-by-step	example
	We decompose a fraction of the form #\frac{\purple l \cdot x+\orange m}{x^2+bx+c}# in which #x^2+bx+c# can be written as #(x+\blue p)^2#.	#\frac{x+1}{x^2+6x+9}#
Step 1	Write #x^2+bx+c=(x+\blue p)^2#.	#x^2+6x+9=(x+\blue 3)^2#
Step 2	Write #\frac{\purple l \cdot x+\orange m}{x^2+bx+c}=\frac{A}{x+\blue p}+\frac{B}{(x+\blue p)^2}#	#\frac{x+\orange1}{x^2+6x+9}=\frac{A}{x+\blue3}+\frac{B}{(x+\blue3)^2}#
Step 3	Bring the right-hand side together in one denominator and compare the numerators on both sides. This gives the equation: \[\purple l \cdot x+\orange m=A\cdot(x+\blue p)+B \]	\[x+\orange1=A \cdot (x+\blue3) + B\]
Step 4	Simplify the equation from step 3 by expanding the brackets, and combining the terms with #x#.	\[x+\orange1=A \cdot x+\blue3A+B\]
Step 5	Compose a system of linear equations by comparing the coefficients of the left and right terms of the equation in step 4.	\[\lineqs{A&=&\purple1\cr \blue3A+B&=&\orange1\cr}\]
Step 6	Solve the system from Step 5 using substitution or elimination.	\[\lineqs{A=1 \cr B=-2}\]
Step 7	Substitute the values found in step 6 in the right-hand side of the equation of step 2 in for the answer.	#\frac{1}{x+\blue3}+\frac{-2}{(x+\blue3)^2}#

General

This step-by-step plan describes fraction decomposition of the denominat denominator for a very specific situation, but this method also works when writing the general denominator as a product. In this course we will not go into that.

Discriminant denominator

The requirement that #x^2+bx+c# can be written as #(x+\blue p) \cdot (x+\green q)# is equivalent to the requirement that the discriminant of the quadratic equation #x^2+bx+c=0# is greater than #0#.

In addition, the requirement that #x^2+bx+c# can be written as #(x+\blue p)^2# is equal to the requirement that the discriminant of the quadratic equation #x^2+bx+c=0# is equal to #0#.

Decompose #\frac{4\cdot x+4}{x^2-15\cdot x+54}# into a sum of two fractions with linear denominators.

#\frac{4\cdot x+4}{x^2-15\cdot x+54}=# #\frac{-{{28}\over{3}}}{x-6}+\frac{{{40}\over{3}}}{x-9}#

Step 1	Using factorization, we find # x^2-15\cdot x+54 = (x-6) \cdot (x-9) #.
Step 2	We write #\frac{4\cdot x+4}{x^2-15\cdot x+54}=\frac{A}{x-6}+\frac{B}{x-9}#
Step 3	We bring the fractions on the right-hand side of the equation together under a common denominator, and compare the numerators. This gives the equation: \[4\cdot x+4=B\cdot \left(x-6\right)+A\cdot \left(x-9\right)\]
Step 4	We simplify the equation from step 3 by expanding the brackets: \[4\cdot x+4=B\cdot x+A\cdot x-6\cdot B-9\cdot A\]
Step 5	We compose a system of linear equations by comparing the coefficients of the equations in step 4. \[\lineqs{B+A=4 \cr -6\cdot B-9\cdot A=4 \cr}\]
Step 6	The solutions of the system from step 5 are equal to: \[\lineqs{A=-{{28}\over{3}} \cr B={{40}\over{3}} \cr}\]
Step 7	We now substitute the values from step 6 in step 2. That gives the answer: \[\frac{-{{28}\over{3}}}{x-6}+\frac{{{40}\over{3}}}{x-9}\]

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