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.
\[\begin{array}{rcl}\dfrac{7x+31}{(x+3) \cdot (x+5)}&=&\dfrac{5}{x+3}+\dfrac{2}{x+5} \\ \\ \dfrac{6x4}{x\cdot (x2)}&=&\dfrac{2}{x}+\dfrac{4}{x2} \\ \\ \dfrac{12x+15}{x^2+x2}&=& \dfrac{9}{x1}+\dfrac{3}{x+2}\end{array}\]
When we look closely at these examples, we see that if we multiply the denominators on the righthand side, we get the denominator on the lefthand 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)#.
Stepbystep fraction decomposition
Stepbystep 
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 righthand 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 righthand side of the equation of step 2 in for the answer. 
#\frac{1}{x+\blue2}+\frac{1}{x+\green3}# 
Step 1  Using factorization, we find # x^211\cdot x+28 = (x4) \cdot (x7) #. 
Step 2  We write #\frac{42\cdot x}{x^211\cdot x+28}=\frac{A}{x4}+\frac{B}{x7}# 
Step 3  We bring the fractions on the righthand side of the equation together under a common denominator, and compare the numerators. This gives the equation: \[42\cdot x=B\cdot \left(x4\right)+A\cdot \left(x7\right)\] 
Step 4  We simplify the equation from step 3 by expanding the brackets: \[42\cdot x=B\cdot x+A\cdot x4\cdot B7\cdot A\] 
Step 5  We compose a system of linear equations by comparing the coefficients of the equations in step 4. \[\lineqs{B+A=2 \cr 4\cdot B7\cdot A=4 \cr}\] 
Step 6  The solutions of the system from step 5 are equal to: \[\lineqs{A={{4}\over{3}} \cr B={{10}\over{3}} \cr}\] 
Step 7  We now substitute the values from step 6 in step 2. That gives the answer: \[\frac{{{4}\over{3}}}{x4}+\frac{{{10}\over{3}}}{x7}\] 
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