- Partial Fractions

We are familiar with the process of combining algebraic fractions, however, in this section we look at the reverse. Partial Fractions decomposition looks at taking a single algebraic fraction and decomposing it into the sum of simpler fractions. This greatly aids in the integration of algebraic functions.


DENOMINATOR WITH LINEAR FACTORS
LESSON 1
Express \(\displaystyle {2x-7\over x^2-x-2}\) in partial fractions. Hence, determine
\[\int {2x-7\over x^2-x-2}\: dx\]
SOLUTION
  • Factorise the denominator.
\(\displaystyle {2x-7\over x^2-x-2}={2x-7\over (x-2)(x+1)}\)
  • Separate the denominator into its linear factors. Each denominator with a linear factor denominator has a numerator that is a constant.
\(\displaystyle {2x-7\over (x-2)(x+1)} ≡{A\over x-2}+{B\over x+1}\)
  • Combine fractions on the right-hand side.
\(\displaystyle {2x-7\over (x-2)(x+1)} ={A(x+1)+B(x-2)\over (x-2)(x+1)}\)
  • Equate numerators.
\(2x-7=A(x+1)+B(x-2)\)
  • Determine the unknown constants by substituting values of \(x\) into both sides of the equation.
  • We should be able to see that when we substitute \(x=-1\) into the equation \(x+1\) will become zero and we can therefore determine the value of \(B\). Additionally substituting \(x=2\) allows us to determine the value of \(A\).
When \(x=-1\)
\(2(-1)-7=A(-1+1)+B(-1-2)\)
\(-9=-3B\)
\(3=B\)
When \(x=2\)
\(2(2)-7=A(2+1)+B(2-2)\)
\(-3=3A\)
\(-1=A\)
  • Substitute the values of \(A\) and \(B\)
\(\displaystyle {2x-7\over x^2-x-2}=-{1\over x-2}+{3\over x+1}\)
  • Now that we have decomposed the original rational expression into partial fractions we can integrate the individual parts.
\(\displaystyle \int {2x-7\over x^2-x-2}\:dx=\int -{1\over x-2}\: dx+\int {3\over x+1}\:dx\)
\(=-\ln⁡|x-2|+3 \ln⁡|x+1|+c\)
\(\displaystyle =\ln⁡|{(x+1)^3\over x-2}|+c\)

DENOMINATOR WITH UNFACTORIZABLE QUADRATIC FACTOR.
LESSON 1
Express \(\displaystyle {x^2+5x+4\over (x^2+3x+1)(x+3)}\) in partial fractions.
SOLUTION
  • When we have an unfactorisable factor in the denominator its numerator is of the form \(Ax+B\).
  • Write each factor of the denominator as the denominator of an individual fraction.
\(\displaystyle {x^2+5x+4\over (x^2+3x+1)(x+3)}={Ax+B\over x^2+3x+1}+{C\over x+3}\)
  • Combine fractions of the right-hand side.
\(\displaystyle {x^2+5x+4\over (x^2+3x+1)(x+3)}={(Ax+B)(x+3)+C(x^2+3x+1)\over (x^2+3x+1)(x+3)}\)
  • Equate numerators.
\(x^2+5x+4=(Ax+B)(x+3)+C(x^2+3x+1)\)
\(=(A+C) x^2+(B+3A+3C)x+3B+C\)
  • Determine the values of the unknowns.
When \(x=-3\)
\(-2=C\)
When \(x=0\)
\(4=3B+C\)
\(4=3B-2\)
\(2=B\)
  • We no longer have any values of \(x\) to substitute into the equation so we will look at equating coefficients
\(1x^2+5x+4=(Ax+B)(x+3)+C(x^2+3x+1)\)
\(=Ax^2+3Ax+Bx+3B+Cx^2+3Cx+C\)
\(=(A+C) x^2+(3A+B+3C)x+3B+C\)
  • Equate coefficients of \(x^2\):
\(A+C=1\)
\(A-2=1\)
\(A=3\)
  • Substitute the determined values of the unknowns.
\(\displaystyle {x^2+5x+4\over (x^2+3x+1)(x+3)}={3x+2\over (x^2+3x+1)}-{2\over x+3}\)

LESSON 2
Express
\(\displaystyle {2x^2-5x+2\over x^3+x}\)
in partial fractions and hence determine
\(\displaystyle \int {2x^2-5x+2\over x^3+x}\:dx\)
SOLUTION
  • Determine the factors of the denominator
\(\displaystyle {2x^2-5x+2\over x^3+x}={2x^2-5x+2\over x(x^2+1)}\)
  • Write the appropriate form of each fraction.
\(\displaystyle {2x^2-5x+2\over x(x^2+1)}={A\over x}+{Bx+C\over x^2+1}\)
  • Combine fractions of the right-hand side.
\(\displaystyle {2x^2-5x+2\over x(x^2+1)}={A(x^2+1)+(Bx+C)(x)\over x(x^2+1)}\)
  • Equate numerators
\(2x^2-5x+2=A(x^2+1)+(Bx+C)x\)
\(=Ax^2+A+Bx^2+Cx\)
\(=(A+B) x^2+Cx+A\)
  • Determine the values of the unknowns.
  • Equate constants:
\(A=2\)
  • NB: The value of \(A\) could have been determined by substituting \(x=0\) into the equation.
  • Equate coefficients of \(x\):
\(C=-5\)
  • Equate coefficients of \(x^2\):
\(A+B=2\)
\(2+B=2\)
\(B=0\)
  • Substitute values for the determined constants.
\(\displaystyle {2x^2-5x+2\over x^3+x}={2\over x}-{5\over 1+x^2}\)
  • Integrate
\(\displaystyle \int {2x^2-5x+2\over x^3+x}\:dx=\int {2\over x}-{5\over 1+x^2}\:dx\)
\(=\ln⁡ x^2-5 \tan^{-1}⁡x+c\)

DENOMINATOR WITH A REPEATED FACTOR
LESSON 1
Express
\(\displaystyle {x^2-3x-9\over x^3-6x^2+9x}\)
in partial fractions
SOLUTION
  • Factorise the denominator.
\(\displaystyle {x^2-3x-9\over x^3-6x^2+9x}={x^2-3x-9\over x(x^2-6x+9)}={x^2-3x-9\over x(x-3)^2}\)
  • Since \((x-3)^2\) is a repeated linear factor we form fractions that include increasing powers of this factor.
\(\displaystyle {x^2-3x-9\over x(x-3)^2}={A\over x}+{B\over x-3}+{C\over (x-3)^2}\)
  • Combine fractions of the left-hand side.
\(\displaystyle {x^2-3x-9\over x(x-3)^2}={A(x-3)^2+Bx(x-3)+Cx \over x(x-3)^2}\)
  • Equate numerators.
\(x^2-3x-9=A(x-3)^2+Bx(x-3)+Cx\)
\(x^2-3x-9=(A+B) x^2+(-6A-3B+C)x+9A-3B\)
  • Determine the values of the unknown constants.
When \(x=3\)
\(-9=3C\)
\(-3=C\)
When \(x=0\)
\(-9=9A\)
\(-1=A\)
  • Equate coefficients of \(x^2\):
\(1=A+B\)
\(2=B\)
  • Substitute the values for each constant.
\(\displaystyle {x^2-3x-9\over x^3-6x^2+9x}=-{1\over x}+{2\over x-3}-{3\over (x-3)^2}\)

IMPROPER FRACTIONS
Rational functions \(\displaystyle {P(x)\over Q(x)}\) are improper if the degree of \(P(x)≥Q(x)\).
LESSON 1
Express \(\displaystyle {2x^3+3x^2-x-4\over x^2 (x+1)}\) in partial fractions and hence determine
\(\displaystyle \int {2x^3+3x^2-x-4\over x^2(x+1)}\:dx\)
SOLUTION
  • We will refer to \(n\) as the degree of the numerator and m as the degree of the denominator. \(n=3\) and \(m=3\). Since \(n-m=0\) we include a constant function, which has degree 0, in addition to the regular partial fraction’s decomposition discussed in this section.
  • Use the appropriate partial fractions decomposition given the terms of the denominator.
\(\displaystyle {2x^3+3x^2-x-4\over x^2(x+1)}=A+{B\over x}+{C\over x^2} +{D\over x+1}\)
  • Combine fractions of the right-hand side.
\(\displaystyle {2x^3+3x^2-x-4\over x^2 (x+1)}={Ax^2 (x+1)+Bx(x+1)+C(c+1)+Dx\over x^2 (x+1)}\)
  • Equate numerators.
\(2x^3+3x^2-x-4=A(x^2 )(x+1)+Bx(x+1)+C(x+1)+Dx^2\)
\(=Ax^3+(A+B+D) x^2+(B+C)x+C\)
  • Determine the values of the unknown constants.
  • Equate coefficients of \(x^3\):
\(A=2\)
  • Equate constants:
\(C=-4\)
  • Equate coefficients of \(x\):
\(B+C=-1\)
\(B-4=-1\)
\(B=3\)
  • Equate coefficients of \(x^2\):
\(A+B+D=3\)
\(2+3+D=3\)
\(D=-2\)
  • Substitute determined values
\(\displaystyle {2x^3+3x^2-x-4\over x^2 (x+1)}=2+{3\over x}-{4\over x^2}-{2\over x+1}\)
  • Integrate
\(\displaystyle \int {2x^3+3x^2-x-4\over x^2(x+1)}\:dx=\int 2\:dx+\int {3\over x}\:dx-\int 4x^{-2}\:dx-\int {2\over x+1}\:dx\)
\(\displaystyle =2x+\ln ⁡x+{4\over x}-2\ln⁡|x+1|+c\)

LESSON 2
Express \(\displaystyle {3x^2+2\over (2x+1)(x-2)}\) in partial fractions and hence determine
\[\int {3x^2+2\over (2x+1)(x-2)}\:dx\]
SOLUTION
  • The degree of the denominator is 2 while the degree of the numerator is 2. Therefore, we have an IMPROPER FRACTION.
  • Express as partial fractions using the appropriate decomposition,
\(\displaystyle {3x^2+2\over (2x+1)(x-2)}=A+{B\over 2x+1}+{C\over x-2}\)
  • Combine the fractions on the right – hand side
\(\displaystyle {3x^2+2\over (2x+1)(x-2)}={A(2x+1)(x-2)+B(x-2)+C(2x+1)\over(2x+1)(x-2)}\)
  • Equate numerators
\(3x^2+2=A(2x+1)(x-2)+B(x-2)+C(2x+1)\)
\(3x^2+2=2Ax^2+(-3A+B+2C)x+(-2A-2B+C)\)
When \(x=2\)
\(14=5C\)
\(\displaystyle {14\over5}=C\)
When \(\displaystyle x=-{1\over2}\)
\(\displaystyle {11\over4}=-{5\over2}B\)
\(\displaystyle -{11\over10}=B\)
  • Equate coefficients of \(x^2\):
\(3=2A\)
\(\displaystyle {3\over2}=A\)
  • Substitute the determined values
\(\displaystyle {3x^2+2\over (2x+1)(x-2)}={3\over2}-{11\over10}(2x+1) +{14\over5}(x-2)\)
  • Integrate
\(\displaystyle \int {3x^2+2\over (2x+1)(x-2)}\:dx=\int {3\over2} \:dx-{11\over20}\int {2\over 2x+1}\:dx+{14\over5}\int {1\over x-2}\:dx\)
\(\displaystyle ={3x\over2}-{11\over20}\ln⁡|2x+1|+{14\over5}\ln⁡|x-2|+c\)

LESSON 3
Express \(\displaystyle {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}\) in partial fractions and hence determine
\[\int {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}\:dx\]
SOLUTION
  • \(n=4\) and \(m=3\). Since \(n-m=1\) we include a linear function in addition to the regular partial fraction’s decomposition discussed in this section.
  • Use the appropriate partial fractions decomposition.
\(\displaystyle {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)} =Ax+B+{C\over x+3}+{D\over x+2}+{E\over x+1}\)
  • Combine fractions on the right-hand side
\(\displaystyle {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}\)
\(\displaystyle ={(Ax+B)(x+3)(x+2)(x+1)+C(x+2)(x+1)+D(x+3)(x+1)+E(x+3)(x+2)\over (x+3)(x+2)(x+1)}\)
  • Equate numerators.
\(x^4+x^3-19x^2-44x-21\)
\(=(Ax+B)(x+3)(x+2)(x+1)+C(x+2)(x+1)+D(x+3)(x+1)+E(x+3)(x+2)\)
  • Determine the values of the unknown constants.
When \(x=-3\)
\(-6=2C\)
\(-3=C\)
When \(x=-2\)
\(-1=-D\)
\(1=D\)
When \(x=-1\)
\(4=2E\)
\(2=E\)
Equate coefficients of \(x^4\):
\(1=A\)
Equate coefficients of \(x^3\):
\(1=6A+B\)
\(1=6+B\)
\(-5=B\)
  • Substitute determined values
\(\displaystyle {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}=x-5-{3\over x+3}+{1\over x-2}+{2\over x+1}\)
  • Integrate
\(\displaystyle \int {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}\: dx\)
\(\displaystyle =\int (x-5)\:dx-3\int {1\over x+3}\:dx+\int {1\over x-2}\:dx+2\int {1\over x+1}\:dx\)
\(\displaystyle ={x^2\over2}-5x-3\ln⁡|x+3|+\ln⁡|x-2|+2\ln⁡|x+1|+c\)