INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS. Introduction It is possible to integrate any rational function, constructed as the ratio of polynomials by epressing it as a sum of simpler fractions - called partial fractions - which we already know how to integrate. As an eample consider the difference of fractions /( and /( + to a common denominator: ( + ( + 5 + ( + ( + Now if we consider the integral of the right hand side, ( + 5 + ln ln + + C. + In general, any rational function may be written as f( P ( Q( where P and Q are polynomials. Denoting P ( a n n +...a + a + a 0 with a n 0, then the degreee of P is n and we write deg(p n. Assuming the degree of P, deg(p is less than deg(q, we may always epress f as a sum of simple fractions. We say such a rational function is proper. If f is improper so that deg(p deg(q then we first divide Q into P using long division until we determine the remainder R( with deg(r < deg(q, then we may write f as f P ( R( S( + Q( Q( where S and R are polynomials. Let us consider an eample of an improper rational function: Eample.. Find 3 +. Proof. The degree of the numerator is higher than the denominator, so we try long division to see if this will simplify the epression 3 ( + ( + ( + + + 3 3 + + + ln + C
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS After long division we attempt to factor the denominator Q( as much as possible. Any polynomial Q can be factored as a product of linear factors (a + b and irreducible quadratic factors (a + b + c, with b 4ac < 0. For eample if Q( 4 6 this could be factored as Q( ( 4( + 4 ( ( + ( + 4 The net step is to epress the proper rational function R(/Q( as a sum of partial fractions of the form A (a + b i, or A + B (a + b + c j We are assured that this is always possible in principle in thanks to a theorem from algebra (The Fundamental theorem of algebra. This will be eplained in the following four cases.. Case I: The Denominator Q( is a Product of Distinct Linear Factors This assumption implies Q( (a + b (a + b...(a k + b k where no factor is repeated nor are any constant multiples of another. From the partial fraction theorem, there will be constants A, A,..., for which ( for each repeated linear term. Eample.. Evaluate R( Q( A A + +... + a + b a + b a k + b k + 3 +3 Proof. Here the degree of the denominator is larger than that of the numerator, so we do not have to do long division, and instead factor the denominator 3 + 3 ( + 3 ( ( +. The denominator has three distinct linear factors, and so we will try + ( ( + A + B + C + To find the values of A, B and C multiply both sides of this equation by the denominator and epand each term + A( ( + + B( + + C( (A + B + C + (3A + B C A As there is equality between both sides, this implies the coefficients of each power must be equal, this produces a linear system of 3 unknowns in terms of 3 equations A + B + C 3A + B C A Solving this system, we find A, B 5 and C 0. Thus,
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 3 + 3 + 3 [ + 5 ] 0 + ln + ln 0 0 ln + + C There is another approach, that eploits the zeros of each linear term. Let us look at another eample to illustrate this approach Eample.. Find a where a 0 Proof. Factoring the denominator, and applying the partial fractions yields a ( a( + a A a + B + a Multiplying the denominator on both sides we have A( + a + B( a Setting a we find A(a, and so A a, then trying a we find ab, implying B a, and so the integral is a ( a a (ln a ln + a + C + a a As ln(/y ln( ln(y this integral is then a a ln a + a + C 3. Case II: Q( is a Product of Linear Factors, with some Repetition If now a linear factor, (a + b is repeated r times in the factorization of Q(, we must replace the term A /(a + b in equation ( with a different term: ( R( Q( A, A, + a + b (a + b +... + A,r (a + b r + A a + b +... + Eample 3.. Determine 4 +4+ 3 + Proof. Long division gives a k + b k 4 + 4 + 3 + + + 4 3 +. Net we factor the denominator, Q(, by first eploiting an obvious root of, namely Q( 0, we may divide by to determine a lower degree polynomial: 3 + ( ( ( ( ( + ( ( + ( is repeated r times, so we will try the partial fraction decomposition 4 3 + A + B ( + C +
4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Multiplying both sides by the denominator we find 4 A( ( + + B( + + C( Setting gives 4 B, gives 4 4C. Thus B and C. There are no more roots to eploit to solve for A, instead we use a 0 as a simple point to solve for nowing B and C 0 A + B + C and so A B + C. With this the integral can be evaluated 4 + 4 + 3 + [ + + + ( + ] + + ln ln + + K + ln + + K 4. Case III: Q( Contains Irreducible Quadratic Factors, with no Repetitions If Q( contains the factor a + b + c with b 4a c < 0 then we have the following partial fraction decomposition R( A + b A (3 Q( a + +... + + b + c a + b a k + b k There will be one of these for each distinct irreducible quadratic factors. The first term in (3 can be integrated by completing the square and using the formula + a ( (4 a arctan + C a Eample 4.. Find +4 3 +4 Proof. As we can only factor 3 + 4 ( + 4 we have + 4 3 A + 4 + B + C + 4 Multiplying by the denominator on the left hand side + 4 A( + 4 + (B + C (A + B C + 4A equating coefficients and solving the linear system we have A, B, and C and the integral is ( + 4 3 + 4 + + 4 breaking the second term into two parts + 4 3 + 4 + + 4 + 4 ln + ln( + 4 arctan(/ + K
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 5 Eample 4.. Evaluate 4 3+ 4 4+3 Proof. The degree of these polynomials are equal, so we first divide 4 3 + 4 4 + 3 + 4 4 + 3 The quadratic in the denominator is irreducible, since 4 4 4 3 3 < 0. To integrate the given function we now complete the square in the denominator 4 4 + 3 ( +. Using the substitution u, with du and (u + : 4 ( 3 + 4 4 + 3 + 4 4 + 3 + (u + u du + u + 4 u + du + u 4 u + du 4 u + du ( u arctan + C + 8 ln(u + 4 + 8 ln(4 4 + 3 4 arctan ( + K Remark 4.3. This illustrates the general procedure for integrating a partial fraction with an irreducible quadratic in the denominator. By completing the square of the denominator in the sum (3 we make a substitution to bring this term into the form Cu + D u + a C u u + a du + D u + a du The first integral will be a logarithm and the second is epressed in terms of arctan. 5. Case IV: Q( Contains a Repeated Irreducible Quadratic Factor If Q( has the factor (a + b + c r with b 4ac < 0 the we must consider the partial fraction: (5 R( Q( A, + b, (a + b + c + A, + b, (a + b + c +... + A,r + b,r (a + b + c r + A +... +. a + b a k + b k Each of these terms can be integrated using a substitution or by completing the square first. Eample 5.. Write the partial fraction decomposition for 3 + + ( ( + + ( + 3
6 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Proof. 3 + + ( ( + + ( + 3 A + B + C + D + + + E + F + + G + H ( + + I + J ( + 3 Eample 5.. Find + 3 ( + Proof. This may be decomposed as + 3 ( + A + B + C + + D + E ( +. Multiplying by the left hand side denominator and epanding out each polynomial 3 + (A + B 4 + C 3 + (A + B + D + (C + E + A we now have a linear system A + B 0, C, A + B + D, C + E, and A. Solving this A, B, C, D, and E 0 so + 3 ( ( + + + + ( + + + + ( + ln ln( + arctan ( + + K Remark 5.3. While this approach will certainly produce a solution it may not be the easiest, for eample the substitution u ( + 3 allows one to rewrite the integral + du ( + 3 3u 3 ln u + C 3 ln 3 + 3 + C 6. Rationalizing Substitutions There are non-rational functions which can be transformed into rational functions using substitutions. For eample if the integral contains g( then the substitution u g( may help. Eample 6.. Find +4 Proof. Let u + 4 then u + 4, implying u 4 and du. The integral is now + 4 u u 4 udu u ( u 4 du + 4 u du 4
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 7 This integral may now be evaluated by factoring u 4 (u (u + or by using formula (4 with a + 4 4 du + u 4 du u + 8 4 ln u u + + C + 4 + ln + 4 + 4 + + C