Integrating rational functions (Sect. 8.4)

Transcription

1 Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots). Integrating rational functions Remark: The problem is to integrate rational functions f (x) = p m(x) q n (x), where p m (x), q m (x) are polynomials degree m, and n. Solution: It can be proven that (5x 3) (x 2 2x 3) dx. (5x 3) (x 2 2x 3) = 2 x x 3. Then, integration is simple: I = 2 ln x ln x 3 + c. Remark: We now present a method to simplify functions f (x) = p m(x), as additions of functions simpler to integrate. q n (x)

2 Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) r r 2 (Non-repeated roots). p (n )(x). (Repeated roots). (x r ) n p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots). The general case. Polynomial division Remark: Before start any integration, use long division to simplify the rational function: f (x) = p m(x) Remark: Here p m and q m are arbitrary polynomials, while r k is a polynomial with degree less than q n. Verify that 4x 2 7 2x + 3 = 2x x + 3. Solution: 2x x + 3 = (2x 3)(2x + 3) + 2 2x + 3 = 4x x + 3

3 Polynomial division 4x 2 7 2x + 3 dx. Solution: The degree of the polynomial in the numerator is greater or equal the degree of the polynomial in the denominator. In this case it is convenient to do the division: 2x 3 2x + 3 ) 4x 2 7 4x 2 6x I = 6x 7 6x + 9 (2x 3) dx dx 2x + 3 4x 2 7 2x + 3 = 2x x + 3. I = x 2 3x + ln(2x + 3) + c. Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots).

4 The method of partial fractions Remarks: We study rational functions r k (x), with k < n. q n (x) : (5x 3) (x + )(x 3) = 2 (x + ) + 3 (x 3). The method is called of partial fractions because the denominators on the right-hand side above contain only part of the denominator on the left-hand side. We present the method through examples. We go from simpler to more complicated situations. Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots).

5 The method of partial fractions (Non-repeated roots) (x )(x + 2) dx. Solution: Denote r =, r 2 = 2. Find a and a 2 such that (x ) (x + 2) = a (x ) + a 2 (x + 2) = a (x + 2) + a 2 (x ). a(x + 2) + b(x ) =. (x )(x + 2) To find a evaluate the equation above at the root r =, = a (3) a = 3. To find a 2 evaluate the equation above at the root r 2 = 2, = a 2 ( 3) a 2 = 3. The method of partial fractions (Non-repeated roots) (x )(x + 2) dx. Solution: Recall: (x ) (x + 2) = a (x ) + a 2 (x + 2), with a = 3, a 2 =. The integral is now simple to evaluate, 3 I = We conclude that (x )(x 2) dx = 3 (x ) dx 3 (x + 2) dx I = 3 ln x 3 ln x c.

6 The method of partial fractions (Non-repeated roots) (x ) (x 2 x 2) dx. Solution: First, find the zeros of the denominator, x 2 x 2 = 0 x ± = [ ] ± (x ) Therefore, we rewrite: I = (x 2)(x + ) dx. { x+ = 2, Partial fraction problem: Find constants a and a 2 such that x =, (x ) (x 2)(x + ) = a (x 2) + a 2 (x + ), r = 2, r 2 =. Do the addition on the right-hand side above: (x ) (x 2)(x + ) = a (x + ) + a 2 (x 2). (x 2)(x + ) The method of partial fractions (Non-repeated roots) Solution: Recall: (x ) (x 2 x 2) dx. The equation above implies: (x ) (x 2)(x + ) = a (x + ) + a 2 (x 2). (x 2)(x + ) x = a (x + ) + a 2 (x 2) To find a evaluate the equation above at the root r = 2, = a (3) a = 3. To find a 2 evaluate the equation above at the root r 2 =, 2 = a 2 ( 3) a 2 = 2 3. We obtain (x ) (x 2)(x + ) = 3 (x 2) (x + ).

7 The method of partial fractions (Non-repeated roots) (x ) (x 2 x 2) dx. Solution: Recall: (x ) (x 2)(x + ) = 3 (x 2) (x + ). The integral is now simple to evaluate, (x ) I = (x 2 x 2) dx = 3 We conclude that (x 2) dx (x + ) dx I = 3 ln x ln x + + c. The method of partial fractions (Non-repeated roots) Theorem (Non-repeated roots - Heaviside cover method) p k (x) The rational function, with n > k and all (x r ) (x r n ) roots r,, r n different, can be written as p k (x) (x r ) (x r n ) = a (x r ) + + a n (x r n ), where the constants a,, a n are given by a = p k (r ) j (r r j ), a n = p k (r n ) j n (r n r j ). Proof: p k (x) = a [ j (x r j) ] + + a n [ j n (x r j) ].

8 Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots). The method of partial fractions (Repeated roots) (2x ) (x 2 6x + 9) dx. Solution: First, find the zeros of the denominator, x 2 6x + 9 = 0 x ± = [ ] 6 ± x ± = 3. Partial fraction problem: Find constants a and a 2 such that (2x ) (x 3) 2 = a (x 3) + a 2 (x 3) 2. Do the addition on the right-hand side above: (2x ) (x 3) 2 = a (x 3) + a 2 (x 3) 2.

9 The method of partial fractions (Repeated roots) (2x ) (x 2 6x + 9) dx. Solution: Recall: (2x ) (x 3) 2 = a (x 3) + a 2 (x 3) 2. Then, 2x = a (x 3) + a 2. To compute a 2 evaluate the expression above at r = 3, 5 = a 2. To compute a derivate the expression above, then evaluate at r = 3, (the evaluation at r = 3 is not needed in this case), 2 = a. We conclude: (2x ) (x 3) 2 = 2 (x 3) + 5 (x 3) 2. The method of partial fractions (Repeated roots) (2x ) (x 2 6x + 9) dx. Solution: Recall: (2x ) (x 3) 2 = 2 (x 3) + 5 (x 3) 2. The integral is now simple to evaluate, (2x ) I = (x 2 6x + 9) dx = 2 (x 3) dx + 5 (x 3) 2 dx We conclude that I = 2 ln x 3 5 (x 3) + c.

10 The method of partial fractions (Repeated roots) Theorem (Repeated roots) The rational function p k(x), with n > k, can be written as (x r) n p k (x) (x r) n = a (x r) + + a n (x r) n, where a i, for i =,, n, is given by a i = p(n i) k (r) (n i)!, Proof: Taking common denominator on the right-hand side above, p k (x) = a (x r) (n ) + a 2 (x r) (n 2) + a (n ) (x r) + a n, a n = p k (r), a (n ) = p (r), a 2 = p(n 2) (r) (n 2)!, a = p(n ) (r) (n )!. Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots).

11 The method of partial fractions (Complex roots) (x + ) 2 (x 2 + ) 2 dx. Solution: Find constants a, b and a 2, b 2 such that (x + ) 2 (x 2 + ) 2 = (a x + b ) (x 2 + ) + (a 2x + b 2 ) (x 2 + ) 2. (x + ) 2 (x 2 + ) 2 = (a x + b )(x 2 + ) + (a 2 x + b 2 ) (x 2 + ) 2, (x + ) 2 = (a x + b )(x 2 + ) + (a 2 x + b 2 ). x 2 + 2x + = a x 3 + a x + b x 2 + b + a 2 x + b 2. x 2 + 2x + = a x 3 + b x 2 + (a + a 2 )x + (b + b 2 ). The method of partial fractions (Complex roots) Solution: Recall: (x + ) 2 (x 2 + ) 2 dx. x 2 + 2x + = a x 3 + b x 2 + (a + a 2 )x + (b + b 2 ). We conclude: a = 0, b =, a 2 = 2, and b 2 = 0. Hence, (a x + b ) I = (x 2 + ) dx + (a2 x + b 2 ) (x 2 + ) 2 dx. I = dx x We conclude that I = arctan(x) 2x dx (x 2 + ) 2. (x 2 + ) + c.

12 The method of partial fractions (Complex roots) Theorem (Repeated roots) p (2n ) (x) The rational function (x + bx + c) n, with b2 4c < 0, can be written as p (2n ) (x) (x 2 + bx + c) n = a x + b (x 2 + bx + c) + + a n x + b n (x 2 + bx + c) n for appropriate constants a i, b for i =,, n. Idea of the Proof: Taking common denominator on the right-hand side above, p (2n ) (x) = (a x + b )(x 2 + bx + c) (n ) + + (a n x + b n ). Expanding the equation above one can find a system of equations for the coefficients. Integrating rational functions (Sect. 8.4) Integrating rational functions, p m(x) q n (x). Polynomial division: p m(x) The method of partial fractions. p (x) (x r )(x r 2 ) p (n )(x). (Repeated roots). (x r ) n The general case. r r 2 (Non-repeated roots). p (2n ) (x) (x 2 + bx + c) n, b2 4c < 0 (Complex roots).

13 The method of partial fractions (General case) Remarks: Consider a general rational function r k(x), with k < n. q n (x) Express the denominator, q, as a product of factors (x r i ) m i and (x 2 + b i x + c i ) l i, with r i roots of q n, and b 2 i 4c i < 0. The partial fraction decomposition for r k q n is the addition of the partial fraction decomposition for each factor in q. The method of partial fractions (General case) 6x 3 8x 2 + 5x 6 (x 2 + )(x 2)x dx. Solution: The partial fraction decomposition is: 6x 3 8x 2 + 5x 6 (x 2 + )(x 2)x = (ax + b) (x 2 + ) + c (x 2) + d x 6x 3 8x 2 +5x 6 = (ax +b)(x 2)x +c(x 2 +)x +d(x 2 +)(x 2) = ax 3 2ax 2 + bx 2 2bx + cx 3 + cx + dx 3 2dx 2 + dx 2d = (a + c + d)x 3 + ( 2a + b 2d)x 2 + ( 2b + c + d)x 2d a + c + d = 6, 2a + b 2d = 8, 5 = 2b + c + d d = 3.

14 The method of partial fractions (General case) Solution: Recall: 9x 3 8x 2 + 5x 6 (x 2 + )(x 2)x dx. a + c + d = 6, 2a + b 2d = 8, 5 = 2b + c + d d = 3. a + c = 3, 2a b = 2, 2b + c = 2. c = 3 a 2b + 3 a = 2 a = 2b 2 4b b = 2. Hence b = 0, and then a = and c = 2. We conclude, 6x 3 8x 2 + 5x 6 [ I = (x 2 + )(x 2)x dx = x (x 2 + ) + 2 (x 2) + 3 x ] dx I = 2 ln(x 2 + ) + 2 ln x ln x + c.