ACTIVITY NO.13 - PARTIAL FRACTIONS Directions: Evaluate each of the integrals listed below. Show your comprehensive solutions dx 1. x2-5x+4 2x+1 2. (x-2) (x-3)2 dx 3. [ dx x2+x+1LESSON 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 (2) D(x) 2 = 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 - ax+b where A is a constant to be determined T64] 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+biz (ax+b)3 ax+bjaWhere: 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 ax +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 (axz + bx + c)", assign a sum of partial fractions A(2ax+b)+B C(2ax+b)+D E(2ax+b) +F K(2ax+b)+L ax'+bx+c (ax'+bx+c)? (ax?+brtej3 + ...+ (ax +bx+c)" Where: A, B, C, D, E, F,...,K,L > constants to be determined Examples: Evaluate the following integrals: X-5 4x2+3x+2 1. J 2-X-2 - dx 2. S x3+x2 -dx 3. x-+3x+5 4. S x5+2x3-3x - dx x3+8 (x2+1)3 Solutions: 1. S- x-5 72-x-2 -dx = S x-5 (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) 747(We need to solve for A and B.) x-5 B (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) (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: A+B=1 Solving for A and B, we have A = - 1 and B = 2 A - 28 = -5 x-5 Hence, J dx (x-2)(x+1) dx = S(+ ) dx = S x+1 = 2 In(x + 1) - In(x -2) + c