Partial Fraction Rational Functions Partial Fraction Finding Partial Fractions Integrating Partial Fraction Eamples
Rational Functions Definition Eample A function of the type P/Q, where both P and Q are polynomials, is a rational function. 3 + 1 + + 1 is a rational function. The degree of the denominator of the above rational function is less than the degree of the numerator. Hence we may rewrite the above rational function in a simpler form by performing polynomial division. Rewriting 3 + 1 = 1+ + + 1 + + 1 For integration, it is always necessary to perform polynomial division first, if possible. To integrate the polynomial part is easy, and one can reduce the problem of integrating a general rational function to a problem of integrating a rational function whose denominator has degree greater than that of the numerator.
Eample of Integrating Rational Functions 1 1 Integrating the functions, and is a simple task 1+ 1+ 1+ applying basic integration formulae and, in the last case, the substitution 1. u = + One gets here we omit the constants of integration 1 1 d = ln 1 +, d = arctan and d = ln 1 + 1+ 1+ 1+ Hence 1+ 1+ 1+ ( ) ( ) 1 1 + + d = ln 1+ + arctan + ln 1 + + K. 3 + 3+ 3 ( ) ( ) ( ) ( ) I.e. d = ln 1+ + arctan + ln 1 + + K. + + + 1 Here K is the constant of integration.
Partial Fraction Decomposition (1) The integration 3 + 3+ 3 ( ) ( ) d = ln 1+ + arctan + ln 1+ + C + + + 1 was based on the decomposition + + + + + 1 1+ 1+ 1+ 3 3 1 1 = + + 3 of the function to be integrated. Definition The above decomposition of the rational function (3 +3+)/( 3 + ++1) is a partial fraction decomposition. Partial Fraction Decomposition is a rewriting of a rational function into a sum a rational functions with as simple denominators as possible. General partial fraction decomposition is technically complicated and involves several cases. It all starts with a factorization of the denominator. The type of the partial factor decomposition depends on the type of the factors of the denominator. The different cases will be eplained on the following slides.
Partial Fraction () The partial fraction decomposition of a rational function R=P/Q, with deg(p) < deg(q), depends on the factors of the denominator Q. Since we are factoring over real numbers, the denominator Q may have following types of factors: 1. Simple, non-repeated linear factors a + b.. Repeated linear factors of the form (a + b) k, k > 1. 3. Simple, non-repeated quadratic factors of the type a + b + c. Since we assume that these factors cannot anymore be factorized, we have b 4 ac < 0. 4. Repeated quadratic factors (a + b + c) k, k>1. Also in this case we have b 4 ac < 0. Finding the partial fraction decompositions is, in all of the above cases, same type of computation. Eventually one wants to integrate the resulting partial fraction decomposition. The integration can, in all cases, be based on formulae, but the computations will get technically complicated.
Simple Linear Factors (1) Case I Consider a rational function of the type P Q ( ) ( ) P( ) ( + )( + ) ( + ) = a b a b a b 1 1 b b i j where aj 0 for all j, for i j, and deg( P) < n = deg( Q ). a a Partial Fraction Decomposition: Case I P( ) ( )( ) ( ) i j A1 A An = + + + a + b a + b a + b a + b a + b a + b 1 1 n n 1 1 n n for some uniquely defined numbers A, k = 1,, n. n k n
Simple Linear Factors () Eample Consider the rational function =. 1 1 + 1 ( )( ) By the result concerning Case I we can find numbers A and B such that A B = = +. To get the 1 ( 1)( + 1) 1 + 1 equations for A and B we use Compute these numbers in the following way the fact that two A B A( + 1) B( 1) polynomials are = + = + 1 1 + 1 1 ( 1)( + 1) ( + 1)( 1) the same if and only if their coefficients are the same. 0i + ( A+ B) + ( A B) A+ B = 0 A = 1 =. 1 1 A B = B = 1 1 1 So the partial fraction decomposition is =. 1 1 + 1
Simple Quadratic Factors (1) Case II P( ) ( ) < ( ) Q( ) ( ) Consider a rational function of the type, deg P deg Q. Assume that the denominator Q has a quadratic factor a + b + c. Partial Fraction Decomposition: Case II The quadratic factor a + b + c of the denominator leads to a term A + B of the type in the partial fraction decomposition. a + b + c
Simple Quadratic Factors () Eample 3 3 The rational function = has a term of 3 1 ( 1)( + + 1) A + B the type in its partial fraction decomposition. + + 1 3 A + B C 3 ( A + B)( 1) C( + + 1) = + = + 3 3 1 + + 1 1 1 ( 1)( + + 1) ( + + 1)( 1) 3 ( ) ( ) A + C + C+ B A+ C B = 1 1 3 3 Hence 3 1 + = + + 3 1 1 1 A+ C = 0 C+ B A = 0 C B = 3 A = 1 B =. C = 1 To get these equations use the fact that the coefficients of the two numerators must be the same.
Repeated Linear Factors (1) Case III ( ) ( ) P Consider a rational function of the type, deg( P) < deg( Q ). Q ( ) ( ) Assume that the denominator Q has a repeated linear factor a+ b, k > 1. k Partial Fraction Decomposition: Case III ( a + b) The repeated linear factor of the denominator leads to terms 1 k of the type in the partial fraction k decomposition. A A A + + + a + b a + b a + b ( ) ( ) k
Repeated Linear Factors () Eample + A B C = + + + 1 + 1 1 4 4 4 3 ( + 1) ( )( ) ( ) ( ) ( 1)( + 1) 1 1 1 1 4 4 4 = 3 A + + B + C + + + 1 ( ) ( ) ( 1)( + 1) 4 + 4 4 4 + 4 4 = 3 The rational function has + 1 ( 1)( + 1) A B C a partial fraction decomposition of the type + +. + 1 1 A+ C + B+ C A B+ C 4 + 4 4 = 3 + 1 A = 3 4 + 4 4 3 1 B = = + + 3 + + ( + 1) C = 1 We get. 1 1 1 ( + 1) Equate the coefficients of the numerators. A+ C = 4 B+ C = 4 A B + C = 4
Repeated Quadratic Factors (1) Case IV ( ) ( ) ( ) ( ) P Consider a rational function of the type, deg( P) < deg( Q ). Q Assume that the denominator Q has a repeated quadratic factor a + b + c, k > 1. k Partial Fraction Decomposition: Case IV ( a + b + c) The repeated quadratic factor of the denominator leads to terms of the type in the partial fraction decomposition. k A 1 + B1 A + B Ak+ Bk + + + a + b + c a + b + c a + b + c ( ) ( ) k
Repeated Quadratic Factors () Eample 3 3 + + 1 1 + 1 4 4 + + = 5 4 3 ( )( ) has a partial A 1 + B1 A + B C fraction decomposition of the type + +. + 1 ( + 1) 1 A 1 + B1 A + B C + + = + 1 ( + 1) 1 This can also be computed by ( A 1 + B1)( + 1)( 1) + ( A + B)( 1) + C( + 1) Maple command 1 + 1 convert. ( )( ) Computing in the same way as before one gets A = B = A = C = 1, 1 1 4 + 3 + 1 1 = = + + 5 4 3 and B 0. Hence. + + 1 + 1 1 ( + 1)
Integrating Partial Fraction (1) Partial Fraction Decomposition is the main method to integrate rational functions. After a general partial fraction decomposition one has to deal with integrals of the following types. There are four cases. Two first cases are easy. A A 1. d = ln a + b + K a + b a Here K is the constant of integration. ( a + b) ( a + b) 1 A A. d = + K, l 1. l a 1 l l
Integrating Partial Fraction () In the remaining cases we have to compute integrals of the type: A + B 3. d and a + b + c A + B 4. d, l > 1 ( a + b + c) l In the case 3, observe that since the denominator cannot be factored further we have 4ac-b > 0. By suitable rewritings and substitutions we get: + B A d = a + b + c a + b ( B Ab/ a) arctan A 4ac b = ln + + + + a 4ac b a bc c K Here K denotes the constant of integration.
Integrating Partial Fraction (3) A + B Integrals of the type d, l > 1, are the last case. ( a + b + c) By repeated integrations by parts these integrals can eventually be computed by the formula for the integrals of type 3. l Theorem P All rational functions f = can be integrated by Partial Q Fraction provided that the polynomial Q can be factored.
Integration Algorithm Integration of a rational function f = P/Q, where P and Q are polynomials can be performed as follows. 1. If deg(q) deg(p), perform polynomial division and write P/Q = S + R/Q, where S and R are polynomials with deg(r) < deg(q). Integrate the polynomial S.. Factorize the polynomials Q and R. Cancel the common factors. Perform Partial Fraction Decomposition to the simplified form of R/Q. 3. Integrate the Partial Fraction Decomposition.
Eamples (1) Eample 1 3 Compute. 3 1 d = + + 3 Observe 1 ( 1)( 1). Hence 3 A B+ C = + 1 1 + + 1 3 for some numbers AB, and C. To compute these numbers AB, and Cwe get 3 A ( + + 1) ( B+ C)( 1) = + 3 + + + + 1 ( 1)( 1) ( 1)( 1) 3 ( A + B) + ( A B+ C) + A C = 3 3 1 1 Hence 3 1 + = + + 3 1 1 1 A+ B = 0 A = 1 A B+ C = 0 B = 1. A C 3 = C =
Eamples () Eample 1 (cont d) 3 Compute. 3 1 d By the previous computations we now have 3 1 + d = d d 3 1 1 + + 1 1 + 1 3 1 = ln 1 d d 1 + + + + 1 1 3 1 = ln 1 ln + + 1 d ( ) ( ) + 1/ + 3 / 4 1 + 1 = ln 1 ln( + + 1) 3 arctan K + 3 Substitute u= ++1 in the first remaining integral and rewrite the last integral. This epression is the required substitution to finish the computation.
Eamples (3) Eample 3 3 + Compute d. 1 We can simplify the function to be integrated by performing polynomial division first. This needs to be done whenever possible. We get: 3 + = + 1 1 Partial fraction decomposition for the remaining rational epression leads to 3 + 1 1 = + = + 1 1 1 + 1 Now we can integrate 1+ 1 1 d = + d 1 1 + 1 1 = + ln 1 ln + 1 + K = + ln + K. + 1
Eamples (4) Eample 3 3 + 1 = + ( + ) 3 + Compute d. ( + ) Partial fraction decomposition in this case yields + This is easy to integrate. It is, however, easier to observe that the substitution Leads to the result immediately (, 3 ) ( 3 ) t = + dt = + d 3 ( + ) + dt d = = ln t + K = ln ( + ) + K. t