Math 2414 Section 7 5 Partial Fractions Method of Partial Fractions 1 2 x 2 x 4 it is a straightforward task to find a common Given a function such as denominator and write the equivalent expression f x x 4 2 x 2 3x x 2 x 4 x 2 x 4 f x 3x x 2 x 8 2 The purpose of partial fractions is to reverse this process Given a rational function that is difficult to integrate the method of partial fractions produces an equivalent function that is much easier to integrate partial fraction decomposition method of 3x 3x 1 2 partial fractions 2 x 2 x 4 x 2 x 4 x 2 x 8 rational function Difficult to integrate 3x dx 2 x 2 x 8 Easy to integrate 2 1 dx x 2 x 4 A B Our objective is to write it in the form x 2 x 4 where A and B are constants to be determined This expression is called the partial fraction decomposition of the original function in this case it has two terms one for each factor in the denominator of the original function The constants A and B are determined using the condition that the original function f and its partial fraction decomposition must be equal for all values of x Simple Linear Factors The previous example illustrates the case of simple linear factors this means the denominator x r which appear to the of the original function consists only of linear factors of the form first power and no higher power Here is the general procedure for this case Procedure Partial Fractions with Simple Linear Factors f x p x q x Suppose where p and q are polynomials with no common factors and with the degree of p less than the degree of q Assume that q is the product of simple linear factors The partial fraction decomposition is obtained as follows x r1 x r2 K x rn where r1 K rn Step 1 Factor the denominator q in the form are real numbers Step 2 Partial fraction decomposition Form the partial fraction decomposition by p x A A2 An 1 K q x x r1 x r2 x rn writing Step 3 Clear denominators Multiply both sides of the equation in Step 2 by q x x r1 x r2 K x rn which produces conditions for A1 A2 K An Step 4 Solve for coefficients Match like powers of x in Step 3 to solve for the undetermined coefficients A1 A2 K An Like a fraction a rational function is said to be in reduced form if the numerator and denominator have no common factors and it is said to be proper if the degree of the numerator is less than the degree of the denominator 3 x 2 7 x 2 f x 3 2 x x 2 x and evaluate Find the partial fraction decomposition for f x dx Solving for more than three unknown coefficients in a partial fraction decomposition may be difficult In the case of simple linear factors the Heaviside Method a technique attributed to Oliver Heaviside 1850 1925 may be used In the previous example the partial fraction decomposition led to the equation 3 x 2 7 x 2 A x 1 x 2 Bx x 2 Cx x 1 Because the equation holds for all values of x it must hold for any particular value of x By choosing a value of x that eliminates all but one term on the right side of the equation it is easy to solve for A B and C Repeated Linear Factors For denominators in which factors are raised to integer powers greater than one we must modify the previous procedure x r Suppose the factor m appears in the denominator where m 1 is an integer Then there x r must be a partial fraction for each power of up to and including the mth power For 4 x 2 x 3 example if appears in the denominator then the partial fraction decomposition includes the terms The rest of the partial fraction remains the same Procedure Partial Fractions for Repeated Linear Factors m x r Suppose the repeated linear factor appears in the denominator of a proper rational function in reduced form The partial fraction decomposition has a partial fraction for each x r power of up to and including the mth power that is the partial fraction decomposition contains the sum A3 Am A1 A2 K 2 3 m x r x r x r x r where A1 K Am are constants to be determined m Be careful This gets wacky for x Why 5 x 2 3x 2 x3 2 x2 dx Setup the necessary integral s to evaluate Irreducible Quadratic Factors A polynomial with real coefficients can be written as the product of linear factors of the form x r and irreducible quadratic factors of the form ax 2 bx c where r a b and c are real numbers 1x2 32 14x 2 343 linear factor repeated linear factor 2 1x 44 22 x4 10 x x 1 43 1 4 2 4 3 2 2 irreducible qudratic factor 2 repeated irreducible quadratic factor Procedure Partial Fractions with Simple Irreducible Quadratic Functions 2 Suppose a simple irreducible factor ax bx c appears in the denominator of a proper rational function in reduced form The partial fraction decomposition contains a term of the form Ax B 2 ax bx c where A and B are unknown coefficients to be determined Give the appropriate form of the partial fraction decomposition for x 2 1 x 2 1 x 2 1 x 4 4 x 3 32 x 2 x 2 x 2 4 x 32 x 2 x 8 x 4 2 x 4 Find the partial fraction decomposition for 2 x 4 x 1 x Evaluate 2 2 1 dx x 1 2 x 2 1 Lastly if the integrand is an improper rational fraction you must reduce the rational fraction to a proper rational fraction plus a polynomial by division 2 x3 8 x 2 9 x 1 x 2 4 x 4 dx Find