Integration of Rational Functions by Partial Fractions
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1 Integration of Rational Functions by Partial Fractions Part 2: Integrating Rational Functions
2 Rational Functions Recall that a rational function is the quotient of two polynomials. x + 3 x + 2 x + 2 x x x x x x = x x x + 2 = 6x2 + 5x 2 x 3 + x 2 2x The left side of this equation is easy to integrate. The right side is hard to integrate. 2
3 Rational Functions Suppose p x and q x are polynomials and p x f x = q x. If the degree of p x is less than the degree of q x, then f x is a proper rational function. If the degree of p x is greater than or equal to the degree of q x, then f x is an improper rational function. 3
4 Partial Fraction Decomposition In theory, a polynomial with real coefficients can always be factored into a product of linear and quadratic factors. If a quadratic factor cannot be further decomposed into linear factors, then it is said to be irreducible. It can be proved that any proper rational function is expressible as a sum of terms (called partial fractions) having the form: A BB + C or aa + b k ax 2 + bb + c k. 4
5 Steps to Partial Fraction Decomposition Suppose p x and q x are polynomials and p x f x = q x. Completely factor the denominator q x into linear and irreducible quadratic factors. Collect all repeated factors so that q x is expressed as a product of distinct factors of the form aa + b m and ax 2 + bb + c m where ax 2 + bb + c m is irreducible. 5
6 Steps to Partial Fraction Decomposition (continued) The structure of p x q x is determined as follows: Linear Factors For each factor of the form aa + b m, introduce the m terms A aa + b + A 2 A m aa + b m aa + b where A, A 2,, A m are constants to be determined. 6
7 Steps to Partial Fraction Decomposition (continued) Irreducible Quadratic Factors For each factor of the form ax 2 + bb + c m, introduce the m terms A x + B ax 2 + bb + c + A 2 x + B 2 ax 2 + bb + c A m x + B m ax 2 + bb + c m where A, A 2,, A m, B, B 2,, B m are constants to be determined. 7
8 Example () x 2 x+3 3 x 2 +x+ 2 = A x + + FF + G x 2 + x + + B x 2 + C x HH + I x 2 + x + 2 D x E x (2) 5x+4 x 2 x 2 +4 = A x + B CC + D x2 + x
9 Example 2 Evaluate x 2 + x 2 dx Solution: x 2 + x 2 = x x + 2 9
10 Example 2 (continued) x x + 2 = A x + B x + 2 Now, multiply both sides of the equation by x x + 2 to get = A x B x To solve for A and B, substitute values of x to make the various terms zero. 0
11 Example 2 (continued) = A x B x Setting x = 2 gives us: = A B 2 = 3B 3 = B Setting x = gives us: = A B = 3A 3 = A
12 Example 2 (continued) An alternate way to solve for A and B is: Take the equation = A x B x Equate corresponding coefficients on both sides = AA + 2A + BB B 0x + = A + B x + 2A B A + B = 0 2A B = Solve the system of equations to get A = 3 and B = 3. 2
13 Example 2 (continued) Since x x + 2 = A x + B x + 2 = 3 x + 3 x + 2 we can now say x x + 2 dx = 3 x dx 3 x + 2 dx = 3 ln x ln x C 3 = 3 ln x x C 3
14 Example 3 Evaluate 3x4 + 3x 3 5x 2 + x x 2 + x 2 dx Solution: Since the degree of the numerator is larger than the degree of the denominator, this is an improper rational function! 4
15 Example 3 (continued) When you have an improper rational function, the first thing you need to do is long division of polynomials to rewrite the improper rational function as the sum of a polynomial and a proper rational function. Using long division of polynomials we get: 3x 4 + 3x 3 5x 2 + x x 2 + x 2 = 3x x 2 + x 2 5
16 Example 3 (continued) 3x4 + 3x 3 5x 2 + x x 2 + x 2 = 3x 2 + dx + dx x 2 + x 2 dx = x 3 + x + 3 See Example 2 ln x x C 6
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