Partial Fractions - Calculus II - Lecture Slides, Slides of Calculus

Download Partial Fractions - Calculus II - Lecture Slides and more Slides Calculus in PDF only on Docsity! 6.3 Partial Fractions Docsity.com A function of the type P/Q, where both P and Q are polynomials, is a rational function. Definition 3 2 1 is a rational function. 1 x x x + + + Example The degree of the denominator of the above rational function is less than the degree of the numerator. First we need to rewrite the above rational function in a simpler form by performing polynomial division. 3 2 2 1 21 1 1 x x x x x x + = − + + + + + Rewriting 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 (is called proper rational function). Thus, polynomial division is the first step when integrating rational functions. Rational Functions Docsity.com Different cases of Partial Fraction Decomposition The partial fraction decomposition of a rational function R=P/Q, with deg(P) < deg(Q), depends on the factors of the denominator Q. It may have following types of factors: 1. Simple, non-repeated linear factors ax + b. 2. Repeated linear factors of the form (ax + b)k, k > 1. 3. Simple, non-repeated quadratic factors of the type ax2 + bx + c. Since we assume that these factors cannot anymore be factorized, we have b2 – 4 ac < 0. 4. Repeated quadratic factors (ax2 + bx + c)k, k>1. Also in this case we have b2 – 4 ac < 0. We will consider each of these four cases separately. Docsity.com Simple Linear Factors ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 1 1 2 2 Consider a rational function of the type P P Q where 0 for all , for , and deg P deg Q . n n ji j i j x x x a x b a x b a x b bba j i j n a a = + + + ≠ ≠ ≠ < =  ( ) ( )( ) ( ) 1 2 1 1 2 2 1 1 2 2 P for some uniquely defined numbers , 1, , . n n n n n k x A A A a x b a x b a x b a x b a x b a x b A k n = + + + + + + + + + =    Case I Partial Fraction Decomposition: Case I Docsity.com Simple Linear Factors 2 2 0 10 2 ( ) ( ) . 1 1 2 1 A B Ax A B x A B x x A B B  + = =+ + + − ⇔ = ⇔ ⇔ − − − = = −  ( )( )2 2 2Consider the rational function . 1 1 1x x x = − − + ( )( )2 By the result concerning Case I we can find numbers and such that 2 2 . x 1 1 1 1 1 A B A B x x x x = = + − − + − + Example 2 2 Compute these numbers in the following way 2 2 ( 1) ( 1) x 1 1 1 1 ( 1)( 1) ( 1)( 1) A B A x B x x x x x x x x + − = + ⇔ = + − − + − − + + − 2 2 1 1So the partial fraction decomposition is . x 1 1 1x x = − − − + To get the equations for A and B we use the fact that two polynomials are the same if and only if their coefficients are the same. Docsity.com Repeated Linear Factors ( ) ( ) ( ) ( ) ( ) ( ) P Consider a rational function of the type , deg P deg Q . Q Assume that the denominator Q has a repeated linear factor , 1. k x x x ax b k < + > Case III ( ) ( ) ( ) 1 2 2 The repeated linear factor of the denominator leads to terms of the type in the partial fraction decomposition. k k k ax b A A A ax b ax b ax b + + + + + + +  Partial Fraction Decomposition: Case III Docsity.com Repeated Linear Factors ( ) ( )( ) ( ) ( ) ( )( ) 2 23 2 2 2 2 3 2 4 4 4 1 1 11 1 1 1 1 4 4 4 11 1 x x A B C x x x x xx A x x B x C x x x x x xx x + − = + + ⇔ + − − + −+ + − + − + + + − = + − −− + ( ) ( ) ( )( ) 2 2 2 3 2 2 4 4 4 11 1 A C x B C x A B C x x x x xx x + + + − − + + − ⇔ = + − −− + 4 2 4 4 A C B C A B C + = ⇔ + = − − + = − ( )( ) ( ) 2 2 23 2 2 4 4 4 4 4 4The rational function has 1 1 1 a partial fraction decomposition of the type . 1 11 x x x x x x x x x A B C x xx + − + − = + − − − + + + + −+ Example 3 2 1 A B C = ⇔ =  = ( ) 2 23 2 4 4 4 3 2 1We get . 1 1 11 x x x x x x xx + − = + + + − − + −+ Equate the coefficients of the numerators. Docsity.com Repeated Quadratic Factors ( ) ( ) ( ) ( ) ( ) ( )2 P Consider a rational function of the type , deg P deg Q . Q Assume that the denominator Q has a repeated quadratic factor , 1. k x x x ax bx c k < + + > Case IV ( ) ( ) ( ) 2 1 1 2 2 22 2 2 The repeated quadratic factor of the denominator leads to terms of the type in the partial fraction decomposition. k k k k ax bx c A x B A x B A x B ax bx c ax bx c ax bx c + + + + + + + + + + + + + +  Partial Fraction Decomposition: Case IV Docsity.com Examples 2 3 2 2 To compute these numbers , and we get 3 ( 1) ( )( 1) 1 ( 1)( 1) ( 1)( 1) A B C A x x Bx C x x x x x x x x + + + − = + − − + + + + − 3 3Compute . 1 dx x −∫ 3 2 3 2 Observe 1 ( 1)( 1). Hence 3 for some numbers , and . 1 1 1 x x x x A Bx C A B C x x x x − = − + + + = + − − + + Example 1 0 1 0 1. 3 2 A B A A B C B A C C + = =   ⇔ − + = ⇔ = −   − = = −  2 3 3 3 ( ) ( ) 1 1 A B x A B C x A C x x + + − + + − ⇔ = − − 3 2 Hence 3 1 2 x 1 1 1 x x x x + = − − − + + Docsity.com Examples 3 3Compute . 1 dx x −∫ 2 2 1 2 1 3 1ln 1 2 1 2 1 xx dx dx x x x x + = − − − + + + +∫ ∫ Example 1 (cont’d) ( ) ( ) 2 2 1 3 1ln 1 ln 1 2 2 1/ 2 3 / 4 x x x dx x = − − + + − + +∫ 3 2 By the previous computations we now have 3 1 2 x 1 1 1 xdx dx dx x x x + = − − − + +∫ ∫ ∫ ( )21 2 1ln 1 ln 1 3 arctan2 3 xx x x K+ = − − + + − +    Substitute u=x2+x+1 in the first remaining integral and rewrite the last integral. This expression is the required substitution to finish the computation. Docsity.com Examples 3 2 2 2 2 1 1 x x x x x − + = + − − 3 2 2 2 1 2 1 1 1 1 1 1ln | 1| ln | 1| ln . 2 2 1 x dx x dx x x x x x xx x K K x − +  = + − − − +  − = + − − + + = + + + ∫ ∫ 3 2 2Compute . 1 x x dx x − + −∫Example 2 We can simplify the function to be integrated by performing polynomial division first. This needs to be done whenever possible. We get: 3 2 2 2 2 1 1 1 1 1 1 x x x x x x x x − + = + = + − − − − + Partial fraction decomposition for the remaining rational expression leads to Now we can integrate Docsity.com