Integration by Partial Fractions✅

Integration by Partial Fractions

Introduction to Partial Fractions

• Partial fractions help evaluate integrals of certain rational functions.
• They also assist in evaluating the Laplace transform in h10o5.

Definition: Degree of a Polynomial

• Consider a polynomial: $p(x) = a<em>n x^n + … + a</em>1 x + a<em>0$, where $a</em>n \neq 0$.
• The degree of the polynomial is the largest power of $x$ that appears in $p(x)$, which is $n$.
• Example: The polynomial $3x^3 + x - 1$ has degree 3.

Definition: Proper Rational Function

• A rational function is a ratio of two polynomials: $\frac{P(x)}{Q(x)}$.
• A rational function is proper when the degree of the numerator (P(x)) is less than the degree of the denominator (Q(x)).
• Examples:
  • $\frac{x}{x^2 + 1}$ is proper (degree of numerator is 1, degree of denominator is 2).
  • $\frac{x^3}{x^2 + 1}$ is not proper (degree of numerator is 3, degree of denominator is 2).
  • $\frac{x}{x^2 - 1}$ is proper (degree of numerator is 1, degree of denominator is 2).

Rewriting Proper Rational Functions

• Proper rational functions can be rewritten as simpler rational functions of degree one and two; these are called partial fractions.
• Example: The proper rational function $\frac{5x - 3}{x^2 - 2x - 3}$ can be written as $\frac{3}{x-3} + \frac{2}{x+1}$.
• The goal is to break down the rational function into a sum of simpler rational functions.
• When a rational function is not proper, extra steps (like long division) are needed to make it proper before applying partial fractions.
• This technique is useful for integrating rational functions by breaking them into simpler terms.

Illustrative Example

• Consider the rational function: $\frac{x+1}{x^2 + 2x - 3}$.

• The degree of the numerator (x+1) is 1, and the degree of the denominator ($x^2 + 2x - 3$) is 2, so it is a proper rational function.

• Factorize the denominator: $x^2 + 2x - 3 = (x+3)(x-1)$.

• Rewrite the rational function as a sum of simpler fractions:
  $\frac{x+1}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1}$, where A and B are constants.

• Task: find A and B.

• Add the fractions:
  $\frac{A}{x+3} + \frac{B}{x-1} = \frac{A(x-1) + B(x+3)}{(x+3)(x-1)}$

• Equate numerators:
  $x+1 = A(x-1) + B(x+3)$

• Expand the numerator:
  $x+1 = Ax - A + Bx + 3B$

• Gather terms with x and constants:
  $x+1 = (A+B)x + (-A+3B)$

• Equate coefficients:

  • Coefficient of x: $1 = A + B$
  • Constant term: $1 = -A + 3B$

• Solve the system of equations:

  • $1 = A + B$
  • $1 = -A + 3B$

• Adding the two equations:
  $2 = 4B \implies B = \frac{1}{2}$

• Substitute B back into the first equation:
  $1 = A + \frac{1}{2} \implies A = \frac{1}{2}$

• Therefore, the rational function can be expressed as:
  $\frac{x+1}{x^2 + 2x - 3} = \frac{1/2}{x+3} + \frac{1/2}{x-1} = \frac{1}{2(x+3)} + \frac{1}{2(x-1)}$

Further Notes on Partial Fractions

• More complicated cases can occur, such as terms in the denominator that are not necessarily linear.
• Consider cases with repeated linear terms.

Example 1

• Denominator: $(x+1)^3$
• Partial fraction decomposition:
  $\frac{1}{(x+1)^3} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{(x+1)^3}$

Example 2

• Denominator: $(x+1)^2(x-1)$
• Partial fraction decomposition:
  $\frac{1}{(x+1)^2(x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1}$

Example 3

• Denominator: $x^3(x-1)^2$
• Partial fraction decomposition:
  $\frac{1}{x^3(x-1)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1} + \frac{E}{(x-1)^2}$

Irreducible Quadratic Denominators

• If the denominator contains an irreducible quadratic expression of the form $ax^2 + bx + c$, where it cannot be factored into real numbers, the partial fraction will have a different form.
• These cases are not covered in ten ninety but will be covered in Ang 10 and five.

Improper Rational Functions and Long Division

• When the rational function is not proper, long division needs to be applied.
• Example: $\frac{x^3}{x^2 - 1}$ is not proper because the degree of the numerator is 3 and the degree of the denominator is 2.
• Apply long division:
  $\frac{x^3}{x^2 - 1} = x + \frac{x}{x^2 - 1}$
• Now, $\frac{x}{x^2 - 1}$ is a proper rational function, so we can apply partial fractions to this term.
  • $\frac{x}{x^2-1} = \frac{x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
  • $x = A(x+1) + B(x-1)$
  • $x = (A+B)x + (A-B)$
  • $1 = A+B$
  • $0 = A-B$
  • $A = B$
  • $1 = 2A \implies A = \frac{1}{2}$
  • $B = \frac{1}{2}$
• So, $\frac{x}{x^2 - 1} = \frac{1/2}{x-1} + \frac{1/2}{x+1}$
• Therefore,
  $\frac{x^3}{x^2 - 1} = x + \frac{1/2}{x-1} + \frac{1/2}{x+1}$

Summarizing the Approach

• Assume the rational function is proper.
  1. Factorize the denominator Q(x).
  2. Write the rational function as a sum of partial fractions according to the table below.

3. Equate numerators over a common denominator, multiply out the factors, and collect all the terms involving x and those that don't involve x; then equilibrate the coefficients on the left hand side.
4. Solve the system of equations to find the coefficients A, B, C.
5. Substitute those into the partial fractions.

Using Partial Fractions for Integration

• Revision of integrals:
  • If $f(x) = ln(ax + b)$, then $f'(x) = \frac{a}{ax + b}$.
  • Therefore, $\int \frac{1}{ax + b} dx = \frac{1}{a} ln|ax + b| + C$ (where C is a constant of integration).
• This result helps integrate rational functions after breaking them down into partial fractions.

Example

• Intgerate $\int \frac{5x - 3}{x^2 - 2x - 3} dx$
  • $\frac{5x - 3}{x^2 - 2x - 3} = \frac{3}{x-3} + \frac{2}{x+1}$
  • $\int \frac{5x - 3}{x^2 - 2x - 3} dx = \int (\frac{3}{x-3} + \frac{2}{x+1})dx$
  • $\int (\frac{3}{x-3} + \frac{2}{x+1})dx = 3\int \frac{1}{x-3}dx + 2\int \frac{1}{x+1}dx$
  • $3\int \frac{1}{x-3}dx + 2\int \frac{1}{x+1}dx = 3ln|x-3| + 2ln|x+1| + c$
  • $3ln|x-3| + 2ln|x+1| + c = ln|x-3|^3 + ln(x+1)^2 + lnk = ln[k|(x-3)^3(x+1)^2|]$
• In h1090, we are going to expect to use partial fraction expansion to evaluate antiderivatives of rational functions. So for instance, if you come across such particular integral, can perhaps use a method of substitution
• Also, one can refer to the previous lecture on integration by substitution.
• Contact me you have any questions.