Expressing a Rational Fraction as the sum of its Partial Fractions

Size: px

Start display at page:

Download "Expressing a Rational Fraction as the sum of its Partial Fractions"

Nathaniel Ellis
6 years ago
Views:

1 PARTIAL FRACTIONS Dear Reader An algebraic fraction or a rational fraction can be, often, expressed as the algebraic sum of relatively simpler fractions called partial fractions. The application of partial fractions lies in evaluating the antiderivative of the rational functions, which is dealt with in std. XII. Example 1:, which can be resolved into partial fractions as : We will now explain how to carry out this process. I) A Brief Recap Definition: An algebraic fraction or a rational fraction is a fraction in which the numerator and the denominator are both polynomial expressions. Example 1:, The above fractions are called proper fractions as the degree of the polynomial in the numerator is lower than the degree of the polynomial in the denominator. Now consider the following fractions: Example 2:, In these cases, the degree of the polynomial in numerator is more than or equal to the degree of the polynomial in the denominator. Such fractions are called improper fractions. II) Expressing a Rational Fraction as the sum of its Partial Fractions CASE I We shall consider the case where the denominator is expressed as product of its linear factors. 19

2 Example: Express as the sum of its partial fractions. To express this algebraic fraction as sum of its partial fractions, we would first express the polynomial in the denominator as product of its factors i.e Therefore, We may now write as sum of its components (proper fractions) as, where A and B are non-zero constants to be determined. Now, (i) As the denominators are same, we need to find the values of A and B for which Comparing the coefficients on both sides we get, 4 = 2; 1 3 Solving the equations simultaneously we get, and Substituting in (i), we get, CASE II We will now consider the case where the denominator in the fraction has a repeated linear factor: 20

3 Example: Express as sum of its partial fractions. Solution: We would first express the denominator in terms of product of its factors, if possible. The possible factors of 2are 1 2 Therefore, Therefore, the possible factors of the denominator are: 1, 1, 2. Now we may write our algebraic fraction as the sum of its partial fractions as: determined (i) ; where A, B, C are non-zero constants to be NOTE: If in case the denominator in the above expression was taken to be 1 2, then its factors would have been 1,1, 1 2. and a fourth fraction would be added to the above, where D would be another constant to be determined. From (i) we get As the denominators are same we need to find the values of A and B for which (Note: We can also achieve the above equation by multiplying both sides of the equation by On comparing the coefficients we get, AC0, 2 1,

4 Solving the above equations we get, 1 9, 2 3, 1 9 Substituting in (i) we get, CASE III We will now consider the case in which the denominator is a quadratic term which cannot be factorised further. Example: Express as sum of its partial fractions. Solution: The two denominators of partial fractions would be 1 1.Here, the quadratic polynomial can t be resolved into further linear factors. We will now express the given fraction as the following Note: The numerator of the fraction having the quadratic polynomial is a linear polynomial in x. = > 1 1 Comparing both sides, we get 0 = (i) 22

5 1 = (ii) 0= A Solving (i) and (ii) simultaneously we get, 1 2 ; 1 2 ; 1 2 Therefore, / // CASE IV We will now consider the case of improper fractions. Example: Express as sum of its partial fractions. In the above case we note that the degree of the polynomial in the numerator is more than the degree of the polynomial in the denominator. Degree of the numerator is 3; Degree of the denominator is 2 In such cases where the degree of term in numerator is more than or equal to the degree of the denominator, we need to perform the long division first. Therefore, 21 (i) Now, consider (ii) Therefore,

6 Comparing the coefficients on both sides we get: 4; 2 11 On solving we get, 4; 3 Substituting in (ii) we get: Therefore (i) can be expressed as NOTE: In general, in case of improper partial fractions, for example, the case above, we may split it into partial fractions as: B Then we may solve the above by comparing the coefficients on both the sides. That is to say, that in case of improper fractions we can start writing the expression in the decreasing powers of to represent the quotient after division. In the case mentioned above the degree of numerator is 3 and the degree of denominator is 2 so we start with the decreasing power of starting with as the difference in the degrees of the polynomial in the numerator and the denominator is 1 and then solve the above. Consider more such examples: ; In this case, the degree of the numerator and denominator is 2, which is same so we start with a constant to represent the quotient. Considering yet another case, 24

7 ; In this case, the degree of the numerator is 4 and the degree of the denominator is 2 so we start writing the expression in the decreasing powers of starting with the degree 2. DO IT YOURSELF Express the following as sum of its partial fractions: 1) 2) 3) - 4) - 5) REFERENCES osition 25

MATHEMATICS Lecture. 4 Chapter.8 TECHNIQUES OF INTEGRATION By Dr. Mohammed Ramidh

MATHEMATICS Lecture. 4 Chapter.8 TECHNIQUES OF INTEGRATION By TECHNIQUES OF INTEGRATION OVERVIEW The Fundamental Theorem connects antiderivatives and the definite integral. Evaluating the indefinite integral,