The method of partial fractions

# The method of partial fractions - Integrating Rational...

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Unformatted text preview: Integrating Rational Functions by the Method of Partial Fraction Examples I When the power of the polynomial of the numerator is less than that of the denominator Example 1 8 − 4x dx ∫ ( x 2 + 1)( x 2 − 2 x + 1) Let 8 − 4x r ( x) = 2 ( x + 1)( x 2 − 2 x + 1) The denominator = ( x 2 + 1)( x 2 − 2 x + 1) = ( x 2 + 1)( x − 1) 2 a b cx + d r ( x) = + +2 x − 1 ( x − 1) 2 x +1 a ( x − 1)( x 2 + 1) + b( x 2 + 1) + (cx + d )( x − 1) 2 = ( x 2 + 1)( x 2 − 2 x + 1) a ( x − 1)( x 2 + 1) + b( x 2 + 1) + (cx + d )( x 2 − 2 x + 1) = ( x 2 + 1)( x 2 − 2 x + 1) (a + c) x 3 + (−a + b − 2c + d ) x 2 + (a + c − 2d ) x + (−a + b + d ) = ( x 2 + 1)( x 2 − 2 x + 1) Comparing the original numerator 8 − 4 x and the new numerator (a + c) x 3 + (− a + b − 2c + d ) x 2 + (a + c − 2d ) x + (− a + b + d ) We get , a+c =0 − a + b − 2c + d = 0 a + c − 2d = −4 −a+b+d =8 Solving this system of linear equations, we get , ⇒ a = −4 b=2 c=4 d =2 a b cx + d r ( x) = + +2 2 x − 1 ( x − 1) x +1 −4 2 4x + 2 = + +2 x − 1 ( x − 1) 2 x + 1 Thus, ∫ r ( x)dx −4 2 4x + 2 = ∫[ + +2dx 2 x − 1 ( x − 1) x +1 ( x − 1) −1 = −4 ln x − 1 + 2 + 2 ln( x 2 + 1) + 2 arctan x + c −1 Example 2 dx ∫ x2 − x 1 1 r ( x) = 2 = x − x x( x − 1) Let 1 a b a( x − 1) + bx =+ = x( x − 1) x ( x − 1) x( x − 1) ( a + b) x − a = x( x − 1) Comparing the two numerators, we get , a + b = 0 and − a = 1 Solving this system of linear equations, we get , ⇒ a = −1 and b = 1 Thus, −1 1 r ( x) = + x ( x − 1) Thus, ∫ r ( x)dx −1 1 = ∫[ +dx x ( x − 1) = − ln x + ln x − 1 + c x −1 = ln +c x Example 3 x − 2x +1 ⋅ dx ∫ ( x 2 + 1) 2 2 x2 − 2x +1 ∫ ( x 2 + 1) 2 ⋅ dx x2 − 2x +1 r ( x) = ( x 2 + 1) 2 ax + b cx + d =2 +2 ( x + 1) ( x + 1) 2 The new numerator = (ax + b)( x 2 + 1) + (cx + d ) = ax 3 + bx 2 + (a + c) x + (b + d ) Comparing numerators and Solving , ⇒ a = 0 , b = 1 , a + c = −2 , b + d = 1 ⇒ c = −2 , d = 1− b = 0 r ( x) ⋅ dx ∫ 1 2x = ∫[ 2 −2dx 2 ( x + 1) ( x + 1) −1 ( x + 1) = arctan x − +c −1 1 = arctan x + 2 +c x +1 2 Examples II When the power of the polynomial of the numerator is equal or greater than that of the denominator Example 1 x +2 ⋅ dx ∫ x +1 2 x2 + 2 ∫ x + 1 ⋅ dx Dividing , we get , x2 + 2 3 r ( x) = = x −1+ x +1 x +1 Alternativ method : x 2 + 2 ( x 2 − 1) + 3 3 = = x −1 + x +1 x +1 x +1 Thus, x2 + 2 3 ⋅ dx = ∫ [ x − 1 +dx ∫ x +1 x +1 x2 = − x + 3 ln x + 1 + c 2 Notice We can use the same method used in this example as an alternative way to write the given rational function as a sum of simpler rational functions (partial fractions). Going back to examples. Notice the following: Example2 1 2 x −x 1 = x( x − 1) (1 − x) + x = x( x − 1) 1− x x = + x( x − 1) x( x − 1) 1 1 =− + x ( x − 1) Example3 x − 2x +1 2 2 ( x + 1) 2 ( x + 1) − 2 x = 2 2 ( x + 1) 1 2x =2 +2 2 x + 1 ( x + 1) 2 Question )( Use two methods dx ∫ x2 −1 Example 2 3 x dx ∫ x2 −1 3 x dx ∫ x2 −1 ( x − x) + x r ( x) = 2 = 2 x −1 x −1 x = x+ 2 x −1 2 x 1 2 r ( x)dx = + ln x − 1 ∫ 22 x 3 3 Examples III Sometimes it is easier to find the constants not by solving a system of linear equations but rather by substituting a different appropriate value for x, in each of these equations‫ز‬ Example 1 15 x − 120 ⋅dx ∫ x( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) 15 x − 120 ∫ x( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) ⋅dx 15 x − 120 r ( x) = x( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) a0 a3 a5 a1 a2 a4 =+ + + + + x x −1 x − 2 x − 3 x − 4 x − 5 The new numerator = a0 ( x − 1)( x − 2)( x − 3)( x − 4)( x − 5) + a1 x( x − 2)( x − 3)( x − 4)( x − 5) + a2 x( x − 1)( x − 3)( x − 4)( x − 5) + a3 x( x − 1)( x − 2)( x − 4)( x − 5) + a4 x( x − 1)( x − 2)( x − 3)( x − 5) + a5 x( x − 1)( x − 2)( x − 3)( x − 4) Substitute, x = 0 The new numerator = a0 (0 − 1)(0 − 2)( x − 3)(0 − 4)(0 − 5) + a1 (0)(0 − 2)(0 − 3)(0 − 4)(0 − 5) + a2 (0)(0 − 1)(0 − 3)(0 − 4)(0 − 5) + a3 (0)(0 − 1)(0 − 2)(0 − 4)(0 − 5) + a4 (0)(0 − 1)(0 − 2)(0 − 3)(0 − 5) + a5 (0)(0 − 1)(0 − 2)(0 − 3)(0 − 4) The original numerator = 15(0) − 120 = −120 Thus, a0 (−1)(−2)(−3)(−4)(−5) = −120 − 120 ⇒ a0 = =1 − (2)(3)(4)(5) Substitute x = 1 The new numerator a1 (1)(1 − 2)(1 − 3)(1 − 4)(1 − 5) The original numerator = 15(1) − 120 = −105 Thus, a1 (1)(−1)(−2)(−3)(−4) = −105 − 105 35 ⇒ a1 = =− (−2)(−3)(−4) 8 Substitute x = 2, we get , a2 (2)(1)(−1)(−2)(−3) = 15(2) − 120 − 90 15 ⇒ a2 = = (2)(1)(−1)(−2)(−3) 2 Substitute x = 3, we get , a3 (3)(2)(1)(−1)(−2) = 15(3) − 120 − 75 25 ⇒ a3 = =− (3)(2)(1)(−1)(−2) 4 Substitute x = 4, we get , a4 (4)(3)(2)(1)(−1) = 15(4) − 120 − 60 5 ⇒ a4 = = (4)(3)(2)(1)(−1) 2 Substitute x = 5, we get , a5 (5)(4)(3)(2)(1) = 15(5) − 120 − 45 3 ⇒ a5 = =− (5)(4)(3)(2)(1) 8 Therfore, ∫ r ( x)dx 35 3 15 25 5 − − − 1 = ∫ [ + 8 + 2 + 2 + 2 + 8 ]dx x x −1 x − 2 x − 3 x − 4 x − 5 35 15 25 = ln x − ln x − 1 + ln x − 2 − ln x − 3 8 2 2 5 3 + ln x − 4 − ln x − 5 + c 2 8 ...
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## This note was uploaded on 10/12/2011 for the course MATH 201 taught by Professor Foad during the Spring '11 term at Qatar University.

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