## - 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
$$\displaystyle \int {x^4+x^3-19x^2-44x-21\over (x+3)(x+2)(x+1)}\: dx$$
$$\displaystyle =\int (x-5)\:dx-3\int {1\over x+3}\:dx+\int {1\over x-2}\:dx+2\int {1\over x+1}\:dx$$
$$\displaystyle ={x^2\over2}-5x-3\ln⁡|x+3|+\ln⁡|x-2|+2\ln⁡|x+1|+c$$