Integration of Rational Functions by Partial Fractions

7 Techniques of Integration

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7.4 Integration of Rational Functions by Partial Fractions

  • Copyright Cengage Learning. All rights reserved.

Integration of Rational Functions by Partial Fractions (1 of 2)

  • Objective: Integrate rational functions (a ratio of polynomials).

  • Method: Express a rational function as a sum of simpler fractions, known as partial fractions, which are easier to integrate.

  • Process: Taking fractions to a common denominator enables integration on the right side of the equation.

Integration of Rational Functions by Partial Fractions (2 of 2)

  • Illustrates how to integrate the right-side function using the method developed.

The Method of Partial Fractions

  • Overview:

    • Consider a rational function where P and Q are polynomials.

    • Possible to express f as a sum of simpler fractions when the degree of P is less than the degree of Q.

    • Such rational functions are termed proper.

The Method of Partial Fractions (1 of 14)

  • If we denote the degree of P by n (with a_n ≠ 0), we establish that deg(P) = n.

  • If f is improper (deg(P) ≥ deg(Q)), we must divide Q into P via long division to yield a remainder R(x) such that deg(R) < deg(Q).

The Method of Partial Fractions (2 of 14)

  • Result of Division:

    • We derive S and R (also polynomials) where S represents the integral of the whole part obtained by division and R is the proper polynomial.

  • Example demonstration: Sometimes, this initial division represents the complete solution.

Example 1

  • Task: Find the integral where the degree of the numerator is greater than the degree of the denominator.

  • Procedure: Perform long division first to represent the function correctly.

The Method of Partial Fractions (4 of 14)

  • Next Step: Factor Q(x) maximally.

  • Every polynomial Q can be expressed as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of the form ax² + bx + c).

  • Example Provided:

    • Similar factorization procedure is illustrated with a polynomial denominator.

The Method of Partial Fractions (5 of 14)

  • Third Step: Express the proper rational function as a sum of partial fractions based on a theorem in algebra.

  • Details: Four different cases arise based on the nature of the denominator.

Case I: Distinct Linear Factors

  • Denominator Q(x) is a product of distinct linear factors.

  • Representation involves no factors being repeated.

  • Partial Fraction Theorem: There exist constants A1, A2, …, Ak such that
    f(x)=A<em>1(ax+b)+A</em>2(cx+d)++Ak(mx+n)f(x) = \frac{A<em>1}{(ax+b)} + \frac{A</em>2}{(cx+d)} + … + \frac{A_k}{(mx+n)}

  • Constants determined through example.

Example 2

  • Task: Evaluate and find partial fraction decomposition.

  • Solution Approach:

    • Since the numerator's degree is less than the denominator's, long division is unnecessary.

    • Factor the denominator as needed.

Example 2 – Solution (1 of 4)

  • Workflow: Denominator has three distinct linear factors, leading to its representation as:
    A(x1)+B(x+2)\frac{A}{(x-1)} + \frac{B}{(x+2)}

  • Determine values A, B, C through multiplication by the common denominator, yielding a solvable equation set.

Example 2 – Solution (2 of 4)

  • Expansion of Equation: Expanding and equating powers of x leads to distinct coefficients, generating a systematic equation set:

    1. Coefficient of x²: 2A+B+2C=12A + B + 2C = 1

    2. Coefficient of x: Equations provide balanced structures for solving constants.

Example 2 – Solution (3 of 4)

  • Outcome: Generates a system of linear equations for A, B, and C with solutions extracted through methods of linear algebra.

Example 2 – Solution (4 of 4)

  • Integration: Utilize substitution: letting u = 2x − 1, (du=2dx)(du = 2dx), leads to direct integration of the obtained terms within the context of function integration.

The Method of Partial Fractions (8 of 14)

  • Alternate Approach: Utilize values of x that satisfy the equation for simplification purposes.

  • By setting x = 0 and strategically chosen points, values become feasible, facilitating easier computation of constants A, B, C.

Case II: Repeated Linear Factors

  • For repeated factors, expression expands to accommodate the repetition:
    A<em>1(a</em>1x+b<em>1)+A</em>2(a<em>1x+b</em>1)2+\frac{A<em>1}{(a</em>1x + b<em>1)} + \frac{A</em>2}{(a<em>1x + b</em>1)^2} + …

  • Illustrative expansion follows sequential logical representation as prior functions with singular repetitions.

Example 4

  • Task: Evaluate
    x42x2+4x+13(x3x2x+1)dx.\int \frac{x^4 - 2x^2 + 4x + 1}{3(x^3 - x^2 - x + 1)} dx.

  • Long Division Step: Execute long division.

Example 4 – Solution (1 of 4)

  • Factoring: Realize that x - 1 is a known factor leading to simpler subsequent expressions in steps.

Example 4 – Solution (2 of 4)

  • Partial Fraction Decomposition Formulation: Employ structure based on derived polynomial characteristics:
    A(x1)2+\frac{A}{(x − 1)^2} + …

  • Utilize least common denominator approach.

Example 4 – Solution (3 of 4)

  • System of Equations for Coefficients: Generate a process where coefficients lead to unique variables resembling the earlier steps concluded.

Example 4 – Solution (4 of 4)

  • Results: Establish values A, B, and C, substituting back to evaluate integral proposition for completion.

Case III: Irreducible Quadratic Factors

  • When Q(x) contains irreducible quadratic factors absent of repetitions, extend expression with quadratic structures:
    Ax+B(quadratic)\frac{Ax + B}{(quadratic)}

  • Example of function with irreducible quadratic is noted to show practical application of the method.

Example 6

  • Task: Evaluate a specific integral tied to quadratics requiring initial long division due to the numerator's degree.

Example 6 – Solution (1 of 3)

  • Completing the Square Approach: Tackle irreducible quadratics by transitioning into perfect square completion techniques.

Example 6 – Solution (2 of 3)

  • Utilization of substitution leads to simplified integral forms positioned for resolution through elementary functions.

Example 6 – Solution (3 of 3)

  • Final evaluation uses trigonometric identities to conclude the computation effectively, establishing the final output in manageable terms.

The Method of Partial Fractions (13 of 14)

  • General Procedure: For integrals resembling forms with squares, employ substitutions that facilitate integration against established forms resulting in logarithms and other identities.

Case IV: Repeated Irreducible Quadratic Factor

  • Presentation of polynomial decomposition expands into repetitive forms where substituted functions maintain operand integrity across variables.

Example 8

  • Task: Evaluate a complicated rational function with multiple polynomial features requiring a robust factorization approach for expansion.

Example 8 – Solution (1 of 2)

  • Equate Coefficients: Resolve generated systems to provide definitively the relationship between variables and their constant counterparts.

Example 8 – Solution (2 of 2)

  • Final integrated representations connect integral steps yielding results across variable shifts and derived functions, showcasing completion of complex rational integration.

Rationalizing Substitutions

  • Overview: Some nonrational functions may transform into rational ones via strategic substitutions; notably, expressions in square root contexts.

Example 9

  • Task: Evaluate the integral involving square roots potentially reshaping the nature of the function and leading to rational forms.

Example 9 – Solution

  • Through calculated substitutions, the transformation into simplified forms yields manageable integral tasks or application of direct computational formulas.