Integration by partial fractions

Transcription

1 Roberto s Notes on Integral Calculus Chapter : Integration methods Section 15 Integration by partial fractions with non-repeated quadratic factors What you need to know already: How to use the integration method of partial fractions when the denominator is a product of non-repeated linear factors. What you can learn here: How to use the integration method of partial fractions when the denominator is a product of non-repeated (irreducible) quadratic factors. I hope you will be comforted by the fact that applying the method of partial fractions in the case when the denominator is a product of non-repeated, irreducible quadratic factors is only a small modification of what you saw happening with non-repeated linear factors. Strategy for splitting a proper fraction whose denominator consists of non-repeated irreducible quadratic factors When a rational function in proper form can be written as: f x p p1 apx ap 1x a1x a0 x b1x c1 x brx cr where all quadratic factors are different from each other, then it can be split into a sum of the form: h1x k1 hrx kr f x x b x c x b x c 1 1 for appropriate values of the coefficients h i and k i. In other words, we can split it as in the linear case, but using linear numerators, instead of constants. And how do we get these coefficients? Do we still have smart values available? Unfortunately not, since these factors are irreducible and therefore have no roots. Instead we do the following. r r Integral Calculus Chapter : Integration methods Section 15 Partial fractions with non-repeated quadratic factors Page 1

2 Strategy for finding the partial fractions coefficients in the presence of irreducible quadratic factors To get the needed values of the coefficients we follow the same procedure as with linear factors, but since there are no smart values to use, we use easy values, that is, small and simple values that make computations easy, such as 0, 1, or other values suggested by the particular case. An example is badly needed, so here it is: Example: f x x 1 x 9 x 1 Notice that the denominator is already factored, something that will occur fairly frequently, unless the factoring is extremely simple, so we apply the procedure: We equate the given fraction to a sum of fractions with a single factor in the denominator and a linear numerator. We can also use an easier notation for the constants: x 1 ax b cx d x 9 x 1 x 9 x 1 Next we combine the fractions on the right side and equate the resulting numerators: x 1 ax b x 1 cx d x 9 x 1 a b 10 c d 1 a b 5c 5d x 1 0 a b 10 c d 0 a b 5c 5d x 1 5 a b 13 c d 1 10a 5b 6c 13d These four equations together form a linear system that can be solved by elimination and back substitution (long way) or by using linear algebra methods. With the latter, and the help of a calculator, we find that: Therefore: 1 1 a b ; c d 8 8 x x 9 x 1 x 9 x 1 Now what? How do we integrate the partial fractions? That is another step where things can get complicated. Since we have no smart values, we use some simple, easy values. x 0 1 b 9d Integral Calculus Chapter : Integration methods Section 15 Partial fractions with non-repeated quadratic factors Page

3 Strategy for integrating a single proper fraction with an irreducible quadratic in the denominator If the quadratic denominator includes a first degree term, we must complete the square before setting up the substitution. If the numerator of the fraction consists of the constant only, an arctangent will be obtained as integral, possibly after a simple substitution. If the numerator of the fraction consists of the linear term only, a substitution will allow us to use the reciprocal rule, thus obtaining a logarithm. If the numerator of the fraction includes both terms of the linear function, the fraction may have to be split further, thus involving both a logarithm and an arctangent. Example: Since x 1 x 9 x 1 x , we can write: 8 x 9 x 1 x 9 x x 9x 1 x x 8 9 x 1 By using the substitution 3u x in the first integral, we obtain: tan x tan x c x 3 Example: x 4x5 Integral Calculus Chapter : Integration methods Section 15 Partial fractions with non-repeated quadratic factors Page 3 In order to integrate this one, we first need to place the denominator in completed square form: x 3 x 3 x 3 x 4x 5 x 4x x 1 We now notice that the derivative of the denominator is x x 4, so we adjust the numerator to show this expression and split the resulting fraction: x x 7 x 3 x x x x u x 1, The first integral can be computed with the substitution while the second can be done with the substitution vx. In this way we get two basic integrals leading to (check the details!): x 3 ln x 1 7 arctan x c x 1 As you can see, there can be quite a bit of work involved, but the procedure works. Need I say that practise is needed to become proficient in the method? Well, I just said it, didn t I! And if you don t see enough questions here, look at the next section on the general case, as well as anything you can find in your resources.

4 Summary To split a rational function with non-repeated irreducible quadratic factors in the denominator, we use the same method as for linear factors, but we need to allow for linear numerators. To find the coefficients of the numerators, we must use easy values, since no smart values are available. Common errors to avoid Watch the algebra, as the process can be quite long, and practice lots to gain familiarity. Learning questions for Section I -15 Review questions: 1. Explain, in your own words, how to integrate a proper rational function whose denominator is a product of non-repeated quadratic factors. Memory questions: 1. In the method of partial fractions, what is needed in the numerators of the smaller fractions when the denominator is irreducible quadratic? Compute the integrals proposed in questions 1-4. Computation questions: 7 6x x 1.. x x3 x x3 Integral Calculus Chapter : Integration methods Section 15 Partial fractions with non-repeated quadratic factors Page 4

5 x x x4 x 5 x x 4 x Theory questions: 1. When do we need to complete the square within the method of partial fractions?. Which two non-rational functions may occur in the integral of a rational function? 1. Construct a simple proper rational function whose denominator consists of one irreducible quadratic factor or two such factors multiplied together. Then integrate it. What questions do you have for your instructor? Integral Calculus Chapter : Integration methods Section 15 Partial fractions with non-repeated quadratic factors Page 5

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