Integration ALGEBRAIC FRACTIONS. Graham S McDonald and Silvia C Dalla


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1 Integration ALGEBRAIC FRACTIONS Graham S McDonald and Silvia C Dalla A selfcontained Tutorial Module for practising the integration of algebraic fractions Table of contents Begin Tutorial c 2004
2 1. Theory 2. Exercises 3. Answers 4. Standard integrals 5. Polynomial division 6. Tips on using solutions Full worked solutions Table of contents
3 Section 1: Theory 3 1. Theory The method of partial fractions can be used in the integration of a proper algebraic fraction. This technique allows the integration to be done as a sum of much simpler integrals A proper algebraic fraction is a fraction of two polynomials whose top line is a polynomial of lower degree than the one in the bottom line. Recall that, for a polynomial in x, the degree is the highest power of x. For example x 1 x 2 + 3x + 5 is a proper algebraic fraction because the top line is a polynomial of degree 1 and the bottom line is a polynomial of degree 2. To integrate an improper algebraic fraction, one firstly needs to write the fraction as a sum of proper fractions. This first step can be done by using polynomial division ( PDivision )
4 Section 1: Theory 4 Look out for cases of proper algebraic fractions whose top line is a multiple k of the derivative of the bottom line. Then, the standard integral k g (x) = k ln g(x) + C g(x) can be used (instead of working out partial fractions) Otherwise, the bottom line of a proper algebraic fraction needs to be factorised as far as possible. This allows us to identify the form of each partial fraction needed factor in the bottom line form of partial fraction(s) (ax + b) (ax + b) 2 A ax+b A ax+b + B (ax+b) 2 (ax 2 + bx + c) Ax+B ax 2 +bx+c where A and B are constants to be determined
5 Section 2: Exercises 5 2. Exercises Click on Exercise links for full worked solutions (there are 13 exercises in total) Perform the following integrations: x 2 + 2x + 5 Exercise 1. x x 3 + 4x Exercise 2. x 2 x 2 + 3x + 4 Exercise 3. 2x 2 + 5x + 3 Exercise 4. x + 2 Theory Answers Integrals PDivision Tips
6 Section 2: Exercises 6 4x Exercise 5. x 4 + 2x + 3 Exercise 6. Exercise 7. Exercise 8. Exercise 9. Exercise 10. x x x (x 3)(x + 4) 118 (2x + 5)(x 7) 7 ()(x 2)(x + 3) 2x + 9 (x + 5) 2 Theory Answers Integrals PDivision Tips
7 Section 2: Exercises 7 13x 4 Exercise 11. (3x 2)(2) Exercise 12. Exercise x (x 2) 2 () 3x 2 (x 1)(x 2 + ) Theory Answers Integrals PDivision Tips
8 Section 3: Answers 8 3. Answers x2 + 2x + 5 ln x + C, x2 + 4x + 3 ln x 1 x + C, x2 + 2x + 2 ln + C, 4. x 2 + x + ln x C, 5. ln x 4 + 2x C, ln x2 5 + C, 7. 2 ln x 3 3 ln x C, ln 2x ln x 7 + C, 9. ln + ln x 2 2 ln x C, ln x x+5 + D, ln 3x ln 2 + C, ln x 2 18 x 2 3 ln x+1 + D, 13. ln x 1 +ln x 2 +x+1 + D.
9 Section 4: Standard integrals 9 4. Standard integrals f (x) x n 1 x f(x) f (x) f(x) xn+1 n+1 (n 1) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+1 n+1 (n 1) ln g (x) e x e x a x ax ln a (a > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x 2 cosech x ln tanh x 2 sec x ln sec x + tan x sech x 2 tan 1 e x sec 2 x tan x sech 2 x tanh x cot x ln sin x coth x ln sinh x sin 2 x cos 2 x x 2 x 2 sin 2x 4 sinh 2 x sinh 2x 4 x 2 + sin 2x 4 cosh 2 x sinh 2x 4 + x 2
10 Section 4: Standard integrals 10 f (x) f (x) f (x) f (x) 1 1 a 2 +x 2 a tan 1 x 1 1 a a 2 x 2 2a ln a+x a x (0< x <a) 1 1 (a > 0) x 2 a 2 2a ln x a x+a ( x > a>0) 1 a 2 x 2 sin 1 x a 1 a 2 +x 2 ( a < x < a) 1 x2 a 2 ln ln x+ a 2 +x 2 a x+ x 2 a 2 a (a > 0) (x>a>0) a2 x 2 a 2 2 [ sin 1 ( ) x a a2 +x 2 a 2 2 ] x2 + x a 2 x 2 a a 2 a [ [ sinh 1 ( x a cosh 1 ( x a ) + x a 2 +x 2 a 2 ] ) + x ] x 2 a 2 a 2
11 Section 5: Polynomial division Polynomial division You can use formal long division to simplify an improper algebraic fraction. In this Tutorial, we us another technique (that is sometimes called algebraic juggling ) In each step of the technique, we rewrite the top line in a way that the algebraic fraction can be broken into two separate fractions, where a simplifying cancellation is forced to appear in the first of these two fractions The technique involves rewriting the topline term with the highest power of x using the expression from the bottom line The detail of how the method works is best illustrated with a long example One such example follows on the next page...
12 Section 5: Polynomial division 12 x 3 + 3x 2 2x 1 = x2 () x 2 + 3x 2 2x 1 { the bottom line has been used to write x 3 as x 2 () x 2 } = x2 () + 2x 2 2x 1 = x2 () + 2x2 2x 1 = x 2 + 2x2 2x 1 = x 2 2x() 2x 2x 1 + { writing 2x 2 as 2x() 2x }
13 Section 5: Polynomial division 13 i.e. x 3 + 3x 2 2x 1 = x 2 + 2x() 4x 1 = x 2 + 2x() = x 2 + 2x + 4x 1 + 4x 1 = x 2 4() x + { writing 4x as 4() + 4 } = x 2 + 2x + 4() + 3 = x 2 + 2x + 4() + 3
14 Section 5: Polynomial division 14 i.e. x 3 + 3x 2 2x 1 = x 2 + 2x + = x 2 + 2x 4 + 4() We have now written the original improper algebraic fraction as a sum of terms that do not involve any further improper fractions, and our task is complete!
15 Section 6: Tips on using solutions Tips on using solutions When looking at the THEORY, ANSWERS, INTEGRALS, PDIVISION or TIPS pages, use the Back button (at the bottom of the page) to return to the exercises Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct Try to make less use of the full solutions as you work your way through the Tutorial
16 Solutions to exercises 16 Full worked solutions Exercise 1. x 2 + 2x + 5 x top line is quadratic in x bottom line is linear in x we have an improper algebraic fraction we need simple polynomial division... i.e. x 2 + 2x + 5 = x = = ( x 2 x + 2x x + 5 ) x ( x ) x 1 x x
17 Solutions to exercises 17 x 2 + 2x + 5 i.e. = 1 x 2 x2 + 2x + 5 ln x + C, where C is a constant of integration. Return to Exercise 1
18 Solutions to exercises 18 Exercise 2. x 3 + 4x x 2 top line is cubic in x bottom line is quadratic in x an improper algebraic fraction simple polynomial division... x 3 + 4x x 2 = = = ( x 3 x 2 + 4x2 x 2 + 3x x ) x 2 (x x ) 2 1 x x + x 2
19 Solutions to exercises 19 i.e. x 3 + 4x x 2 = 1 2 x2 + 4x + 3 ln x + x 1 ( 1) + C = 1 2 x2 + 4x + 3 ln x 1 x + C, where C is a constant of integration. Return to Exercise 2
20 Solutions to exercises 20 Exercise 3. x 2 + 3x + 4 top line is quadratic in x bottom line is linear in x an improper algebraic fraction polynomial division... Now we have more than just a single term in the bottom line and we need to do full polynomial division If you are unfamiliar with this technique, there is some extra help within the PDivision section Here, we will go through the polynomial division first, and we will leave the integration until later...
21 Solutions to exercises 21 x 2 + 3x + 4 = x() x + 3x + 4 { the bottom line has been used to write x 2 as x() x } = x() + 2x + 4 = x() + 2x + 4 = x + 2x + 4 2() = x + { writing 2x as 2x() 2 }
22 Solutions to exercises 22 i.e. x 2 + 3x + 4 = x + 2() + 2 = x + x 2 + 3x + 4 = 2() = x { polynomial division is complete, since we no longer have any improper algebraic fractions } ( x ) = 1 2 x2 + 2x + 2 ln + C. Return to Exercise 3
23 Solutions to exercises 23 Exercise 4. 2x 2 + 5x + 3 x + 2 top line is quadratic in x bottom line is linear in x an improper algebraic fraction polynomial division... 2x 2 + 5x + 3 x + 2 = = = 2x(x + 2) 4x + 5x + 3 x + 2 { the bottom line has been used to write 2x 2 as 2x(x + 2) 4x } 2x(x + 2) + x + 3 x + 2 2x(x + 2) + x + 3 x + 2 x + 2
24 Solutions to exercises 24 i.e. 2x 2 + 5x + 3 = 2x + x + 3 x + 2 x + 2 = 2x + (x + 2) x + 2 { writing x as (x + 2) 2 } = 2x + (x + 2) + 1 x + 2 = 2x + (x + 2) 1 + x + 2 x + 2 = x + 2 { no improper algebraic fractions } 2x 2 ( + 5x + 3 = ) x + 2 x + 2 = x 2 + x + ln x C. Return to Exercise 4
25 Solutions to exercises 25 Exercise 5. 4x x 4 + 2x + 3 top line is degree 3 in x bottom line is degree 4 in x we have a proper algebraic fraction factorise bottom line for partial fractions? No! First, check if this is of the form k g (x) g(x), where k =constant If g(x) = x 4 + 2x + 3 (the bottom line), g (x) = dg = 4x3 + 2 (which exactly equals the top line). So we can use the standard integral k g (x) = k ln g(x) + C, with k = 1 g(x) (or employ substitution techniques by setting u = x 4 + 2x + 3) 4x x 4 + 2x + 3 = ln x4 + 2x C. Return to Exercise 5
26 Solutions to exercises 26 Exercise 6. x x 2 5 top line is degree 1 in x bottom line is degree 2 in x we have a proper algebraic fraction consider for partial fractions? No! First, check if this is of the form k g (x) g(x), where k =constant If g(x) = x 2 5 (the bottom line), g (x) = dg = 2x (which is proportional to the top line). So we can use the standard integral k g (x) = k ln g(x) + C, with k = 1 g(x) 2 (or employ substitution techniques by setting u = x 2 5) i.e. x x 2 5 = 1 2 2x x 2 5 = 1 2 ln x2 5 + C. Return to Exercise 6
27 Solutions to exercises 27 Exercise x (x 3)(x + 4) is a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line Try partial fractions 17 x (x 3)(x + 4) = = A x 3 + B x + 4 A(x + 4) + B(x 3) (x 3)(x + 4) 17 x = A(x + 4) + B(x 3) [ if true then true for all x ] x = 4 gives = 0 + ( 4 3)B i.e. 21 = 7B, B = 3 x = 3 gives 17 3 = (3 + 4)A + 0 i.e. 14 = 7A, A = 2
28 Solutions to exercises x (x 3)(x + 4) = 2 x 3 + ( 3) x + 4 = 2 x 3 3 x + 4 = 2 ln x 3 3 ln x C. Note. In the above we have used ax + b = 1 ln ax + b + D a Return to Exercise 7
29 Solutions to exercises 29 Exercise (2x + 5)(x 7) is a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line Try partial fractions 118 (2x + 5)(x 7) = = A 2x B x 7 A(x 7) + B(2x + 5) (2x + 5)(x 7) 118 = A(x 7) + B(2x + 5) x = 7 gives = (14 + 5)B i.e. 95 = 19B, B = 5 x = 5 2 gives = ( 5 2 7)A i.e = 19 2 A, A = 1
30 Solutions to exercises (2x + 5)(x 7) = = 1 2x x 7 2x x 7 = 1 ln 2x ln x 7 + C. 2 Note. In the above we have used ax + b = 1 ln ax + b + D a Return to Exercise 8
31 Solutions to exercises 31 Exercise 9. 7 ()(x 2)(x + 3) is a proper algebraic fraction, and the top line is not a multiple of the derivative of bottom line Try partial fractions 7x+1 (x+1)(x 2)(x+3) = A x+1 + B x 2 + C x+3 = A(x 2)(x+3)+B(x+1)(x+3)+C(x+1)(x 2) (x+1)(x 2)(x+3) 7 = A(x 2)(x + 3) + B()(x + 3) + C()(x 2)
32 Solutions to exercises 32 7 = A(x 2)(x + 3) + B()(x + 3) + C()(x 2) x = 1 gives 6 = A( 3)(2) i.e. 6 = 6A i.e. A = 1 x = 2 gives 15 = B(3)(5) i.e. 15 = 15B i.e. B = 1 x = 3 gives 20 = C( 2)( 5) i.e. 20 = 10C i.e. C = 2 7 ()(x 2)(x + 3) = 1 x x x+3 = ln + ln x 2 2 ln x C. Return to Exercise 9
33 Solutions to exercises 33 Exercise 10. Proper algebraic fraction and we can use partial fractions 2x + 9 (x + 5) 2 = A (x + 5) + B (x + 5) 2 where 2x + 9 A(x + 5) + B = (x + 5) 2 (x + 5) 2 i.e. 2x + 9 = A(x + 5) + B x = 5 gives = B i.e. B = 1 x = 0 gives 9 = 5A + B = 5A 1 i.e. 10 = 5A i.e. A = 2 2x + 9 (x + 5) 2 = 2 x+5 + ( 1) (x+5) 2 = 2 x+5 (x+5) 2
34 Solutions to exercises 34 i.e. 2x + 9 = 2 ln x+5 (x + 5) 2 + C (x + 5) 2 (x + 5) 1 = 2 ln x+5 + C ( 1) = 2 ln x x C, where, in the last integral, we have used (ax + b) n = (ax + b)n+1 n C, (n 1). Return to Exercise 10
35 Solutions to exercises 35 Exercise 11. Proper algebraic fraction and we need to use partial fractions 13x 4 (3x 2)(2) = A (3x 2) + B (2) where 13x 4 A(2) + B(3x 2) = (3x 2)(2) (3x 2)(2) 13x 4 = A(2) + B(3x 2) and x = 1 2 gives = B ( 3 2 2) i.e = 7 2B, i.e. B = 3 x = gives = A ( ) 3 3 i.e = 7 3 A i.e. A = 2
36 Solutions to exercises 36 13x 4 (3x 2)(2) = 2 3x x+1 = 2 3x x+1 ( ) ( 1 1 = 2 ln 3x ) ln 2 + C = 2 3 ln 3x ln 2 + C, 2 where ax + b = 1 ln ax + b + C has been used. a Return to Exercise 11
37 Solutions to exercises 37 Exercise 12. Use Partial fractions 27x (x 2) 2 () = A (x 2) + B (x 2) 2 + C where 27x = A(x 2)() + B() + C(x 2) 2 x = 2 gives 54 = 3B i.e. B = 18 x = 1 gives 27 = C( 3) 2 i.e. C = 3 x = 0 gives 0 = A( 2) ( 3)(4) i.e. A = 3
38 Solutions to exercises 38 27x (x 2) 2 () = 3 x (x 2) 2 3 = 3 ln x (x 2) 2 3 ln x+1 +D = 3 ln x ( 1) (x 2) 1 3 ln x+1 + D = 3 ln x ln x+1 + D. x 2 Return to Exercise 12
39 Solutions to exercises 39 Exercise 13. 3x 2 (x 1)(x 2 + ) = A x 1 + Bx + C x 2 + Note that x 2 + does not give real linear factors One thus uses the partial fraction We then have Bx+C x 2 +x+1 3x 2 = A ( x 2 + ) + (Bx + C)(x 1) x = 1 gives 3 = 3A i.e. A = 1 x = 0 gives 0 = A C i.e. C = A = 1 x = 1 gives 3 = A( ) + ( B + C)( 2)
40 Solutions to exercises 40 i.e. 3 = A + 2B 2C i.e. 3 = 1 + 2B 2 i.e. 4 = 2B i.e. B = 2 3x 2 (x 1)(x 2 + ) = = A x 1 + x 1 + Bx + C (x 2 + ) 2 x 2 + = ln x 1 +ln x 2 +x+1 + D, and we note that the second integral is of the form g (x) = ln g(x) + g(x) DṘeturn to Exercise 13
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