ACTIVITY No.13 PARTIAL FRACTIONS Directions: Evaluate each of the integrals listed below. Show your comgrehensive solutions :12: 1' I x25x+4 2x+1 2- I ma\" LESSON 16: TECHNIQUES - RATIONAL (PARTIAL) FRACTIONS XI. Rational Fractions Objectives: 1. Convert integrand into something that can be solved using previously discussed theorems 2. Make the integration of complicated functions easier 3. Express proper rational fractions as a sum of partial fractions. 4. Integrate partial fractions where the factors of the denominators are linear and/or quadratic. General Forms: N(x) D(x (proper) N(x) R(x) D(x) = Q(x) + D(X) (improper) Where: N = numerator, D = denominator, Q = partial quotient, R = remainder General Rules: 1. Make sure that the fraction is PROPER. 2. Convert proper rational fractions into sum of PARTIAL fractions. 3. Integrate. Conversion of Proper Rational Fractions into Partial Fractions T63] Case 1: The factors of D(x) are all LINEAR and NONE is REPEATED. To each non-repeating linear factor ax + b (or ax only), assign a partial fraction - where A is a constant to be determinedT64] Case 2: The factors of D(x) are all LINEAR and SOME are REPEATED. To each repeating linear factor (ax + b)" or (or x"), assign a sum of partial fractions + ax+b (ax+b) 2 7 (axtb)3+ ... + (ax+b)n Where: A, B, C, ., L are constants to be determined T65] Case 3: D(x) contains irreducible QUADRATIC factors and NONE is REPEATED. To each nonrepeating irreducible quadratic factor ax + bx + c, assign a partial fraction: A(2ax+b)+B ax2+bx+c Where: A, B -> constants to be determined 2ax + b = derivative of the quadratic factor T66] Case 4: D(x) contains irreducible QUADRATIC factors and SOME are REPEATED. To each repeating irreducible quadratic factor (ax? + bx + c)", assign a sum of partial fractions A(2ax+b)+B C(2ax+b)+D E(2ax+b)+F K(2ax+b)+L ax2+bx+c (ax2+bx+c)2 (ax2 +batc)3+ ..+ (ax2+bx+c)" Where: A, B, C, D, E, F,.,K,L > constants to be determinedExamples: Evaluate the following integrals: 1. x-5 -dx 2. 4x-+3x+2 -dx 2-x-2 x3+x2 3. 2+3x+5 4. 5+2x3-3x (x2+1)3 dx x3+8 Solutions: x-5 x-5 1. J 2-x-2 - dx = J (x-2) (x+1) dx (This is Case 1. The given denominator is quadratic but reducible/factorable so it cannot be considered as Case 3.) x-5 Resolve into: (x-2) (x+1) x- 2 x+1 (We need to solve for A and B.) r-5 (x-2) (x+1) x-2 x+1 > Multiply both sides by the LCD (x - 2)(x+ 1) x-5 [(x-2) (x+1) x-2 + x47 ( x - 2 ) (x + 1) x - 5 = A(x + 1) + B(x - 2) > x -5 = Ax+ A+ Bx -2B We use the method of coefficients: For coefficients of x: 1 = A + B (eqn 1) For coefficients of x (constants): -5 =A -2B (eqn 2) We now have a system of linear equations in two variables:x - 5 = A(x + 1) + B(x - 2) > x -5 = Ax+ A+ Bx - 2B We use the method of coefficients: For coefficients of x: 1 = A + B (eqn 1) For coefficients of xo (constants): -5 = A - 2B (eqn 2) We now have a system of linear equations in two variables: A+B =1 Solving for A and B, we have A = - 1 and B = 2 A - 2B = -5 x-5 Hence, J (x-2) (x+1) dx = S (+ )dx = S (+2dx = 2In(x + 1) - In(x - 2) + C