DEPARTMENT OF CIVIL ENGINEERING AND GEOMATICS MATHEMATICS 1 – MTH158S PARTIAL FRACTIONS CHAPTER 68 IN BIRD
Lecturer: A. Cloete
Overview of Partial Fractions
The concept of partial fractions can be illustrated using simple fractions:
\frac{1}{4} + \frac{1}{5} = \frac{9}{20}This can be rewritten as:
\frac{9}{20} = \frac{1}{4} + \frac{1}{5}Here we have resolved \frac{9}{20} into two partial fractions \frac{1}{4} and \frac{1}{5} .
A similar process is applied to algebraic fractions.
Define F(x) = \frac{p(x)}{q(x)} where p(x) and q(x) are polynomials in x .
F(x) is referred to as a rational fraction.
Division of Rational Fractions
When the degree of the numerator is greater than or equal to the degree of the denominator:
Perform polynomial long division.
Example:
\frac{2x^4 + x^2 - 3x + 1}{x^3 - 2x^2 - 3x} = 2x + 4 + \frac{15x^2 + 9x + 1}{x^3 - 2x^2 - 3x}
The first two terms can be integrated directly, but the remaining fraction requires further manipulation.
Breaking Down into Partial Fractions
If the denominator is factorable, the fraction is decomposed into simpler partial fractions.
There are four distinct cases to consider based on the nature of the factors:
Case 1: Linear Non-Repeated Factors
Structure: f(x) = (ax+b)(cx+d)(ex+f)
Decomposition:
f(x) = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{ex+f}
Case 2: Linear Repeated Factors
Structure: f(x) = (ax+b)^n
Decomposition:
f(x) = \frac{A}{(ax+b)^n} + \frac{B}{(ax+b)^{n-1}} + \ldots + \frac{P}{ax+b}
Case 3: Quadratic Non-Repeated Factors
Structure: f(x) = (ax^2+bx+c)(dx^2+ex+f)
Decomposition:
f(x) = \frac{Ax+B}{ax^2+bx+c} + \frac{Cx+D}{dx^2+ex+f}
Case 4: Quadratic Repeated Factors
Structure: f(x) = (ax^2+bx+c)^n
Decomposition:
f(x) = \frac{Ax+B}{(ax^2+bx+c)^n} + rac{Cx+D}{(ax^2+bx+c)^{n-1}}
Steps for Decomposition
Check if the degree of the numerator is greater than or equal to the degree of the denominator.
If so, use long division to simplify.
Check if the fraction is factorable.
If factorable, break it into simpler partial fractions.
Preferentially, linear factors are easier to work with, hence always attempt to factor quadratic functions into linear components if possible.
Depending on the type of factors, apply the suitable decomposition case.
Examples of Decomposition
Example 1: \frac{x-1}{x^3+x^2-12x}
Check:
Is the degree of (x-1) smaller than the degree of (x^3+x^2-12x)?
Yes - No need for long division.
Is the denominator factorable?
Yes - Factor:
x^3+x^2-12x = x(x^2+x-12) = x(x+4)(x-3)
Therefore:
\frac{x-1}{x(x+4)(x-3)} = \frac{A}{x} + \frac{B}{x+4} + \frac{C}{x-3}This leads to:
x-1 = A(x+4)(x-3) + Bx(x-3) + Cx(x+4)
Solving for Coefficients
For x = 0 :
-1 = A(4)(-3) \Rightarrow A = \frac{1}{12}For x = -4 :
-5 = B(-4)(-7) \Rightarrow B = \frac{5}{28}For x = 3 :
2 = C(3)(7) \Rightarrow C = \frac{2}{21}
Result:
\frac{x-1}{x(x+4)(x-3)} = \frac{1}{12}x + \frac{5}{28}(x+4) + \frac{2}{21}(x-3)
Example 2: ( \frac{t^3-t+3}{t^2(t-2)} )
Here, the degrees are equal, necessitating the use of long division.
\frac{t^3-t+3}{t^2(t-2)} = 1 + \frac{2t^2-t+3}{t^2(t-2)}Now, resolve to find:
2t^2-t+3 = A(t-2)+B t(t-2)+C t^2
Find Coefficients
For t = 0 :
3 = A(-2) \Rightarrow A = -\frac{3}{2}For t = 2 :
9 = 4C \Rightarrow C = \frac{9}{4}Substitute and group terms to find B :
Result:
\frac{t^3-t+3}{t^2(t-2)} = 1 - \frac{3}{2}(t-2) - \frac{1}{4}t + \frac{9}{4}(t-2)
Exercises
Break the following fractions into partial fractions:
\frac{3x-7}{(x-1)(x-2)(x-3)}
\frac{4x+3}{4x^3+8x^2+3x}
\frac{1}{(x+1)(x^2+1)}
\frac{1}{x(x+1)^2}
\frac{1}{x(x^2+x+1)}
\frac{x+1}{x^2(x-1)}
\frac{x^3+x^2}{x^2+x-2}
\frac{1}{x^4+4x^2+3}
\frac{1}{(x+3)(x^2-x-2)}
\frac{x^2+x+1}{(x+3)(x-1)(x-2)}
\frac{5x^2-1}{x^3-x}
\frac{4x^3+2x^2+1}{4x^3-x}
\frac{x^2-3}{(x+1)^2(x+2)}
\frac{x^2}{(4x^2-1)(2x+3)}
\frac{2x^2+1}{(x+2)^3}
\frac{x^3}{(x^2+1)^2}
\frac{x^2+x}{x^3-x^2+x-1}
\frac{2}{(x^3+x^2+4)(x^2+4)}
\frac{x^3-3x^2+x-1}{(x^2-4x+4)}
\frac{x^4-x^3-3x^2-2x+1}{(x^3+x^2-2x)}