Study 7.4 # 1 11, 15, 17, 21, 25. Class Notes: Prof. G. Battaly, Westchester Community College, NY. x x 2 +4x+3. How do we integrate?
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1 Goals: 1. Recognize that rational expressions may need to be simplified to be integrable. 2. Use long division to obtain proper fractions, with the degree of the numerator less than the degree of the denominator. 3. Integrands that are rational expressions may be decomposed to simpler forms that are integrable. To decompose a rational expression: a) factor the denominator b) write the problem as the sum of fractions using the factors as denominators c) for numerators, use the polynomial form with a degree lower than the degree of the denominator: A for a linear denominator, Bx+C for a quadratic denom Study 7.4 # 1 11, 15, 17, 21, 25 x x 2 +4x+3 dx How do we integrate? 1. u substitution? need 2x + 4. missing the 2 2. trig substitution? might work if complete square, etc. 3. parts? need factors and an integrable part that gets less complicated when integrated
2 Adding simple fractions results in more complex composed fractions = + = = + = x x+1 x(x+1) x 2 +x x dx x 2 +4x+3 To integrate, can decompose the fraction HOW? examples: 7 = x+3 = 2x+3 = + x 2 +x x(x+1) x x+1??
3 decompose the fraction #1 1. Factor denominator 2. Write as sum of fractions denom are factors of orig denom 2x+3 = 2x+3 = A + B x 2 +x x(x+1) x x+1 The goal is to build a rational expression using the unknown coefficients and then equate the coefficients to those in the given numerator. This results in a decomposition to the original addends. decompose the fraction 1. Factor denominator 2. Write as sum of fractions denom are factors of orig denom 3. For numerators, use degree less than degree of denom 2x+3 = 2x+3 = A + B x 2 +x x(x+1) x x+1 4. Find common denominators and add fractions A + B = A(x+1) + Bx = Ax+A+Bx x x+1 x(x+1) x(x+1) 5. Combine like terms and equate numerators. A, B constants 2x+3 = (A+B)x+A x 2 +x x(x+1) and 2x+3 = (A+B)x+A
4 decompose the fraction denom are factors of orig denom 1. Factor denominator 2. Write as sum of fractions whose denom are factors of the orig denom 3. For numerators, use degree less than denominator 4. Find common denominators and add fractions 5. Combine like terms and equate numerators. 2x+3 = 2x+3 = A + B x 2 +x x(x+1) x x+1 = A(x+1) + Bx = Ax+A+Bx x(x+1) x(x+1) 2x+3 = (A+B)x+A x 2 +x x(x+1) 2x+3 = (A+B)x+A 6. Equate unknowns to coefficients. A + B = 2 A = 3, B = 1 2x+3 = A + B = x 2 +x x x+1 x x+1 x dx x 2 +4x+3 To integrate, can decompose the fraction #2 x = x = A + B x 2 +4x+3 (x+3)(x+1) x+3 x
5 x dx x 2 +4x+3 To integrate, decompose the fraction x = x = A + B x 2 +4x+3 (x+3)(x+1) x+3 x+1 x = (A+B)x+(A+3B) x 2 +4x+3 (x+3)(x+1) = A(x+1) + B(x+3) = Ax+A+Bx+3B (x+3)(x+1) (x+3)(x+1) x = (A+B)x + (A+3B) 1x+0 = (A+B)x + (A+3B) A + B = 1 A +3B= 0 2B= 1, B= 1/2, A=3/2 x = A + B = x 2 +4x+3 x+3 x+1 2(x+3) 2(x+1) x dx x 2 +4x+3 To integrate, decompose the fraction x = A + B = x 2 +4x+3 x+3 x+1 2(x+3) 2(x+1) x dx = 3 dx 1 dx x 2 +4x+3 2 x +3 2 x+1 = 3 ln x+3 1 ln x+1 + c
6 x 1 dx x 1 = A + B + C + D x 3 (x+1) x 3 (x+1) x+1 x x 2 x 3 #3 x 1 dx x 1 = A + B + C + D x 3 (x+1) x 3 (x+1) x+1 x x 2 x 3 #3 Do we really need x, x 2, and x 3? x 1 = Ax 3 + Bx 2 (x+1) + Cx(x+1) + D(x+1) x 1 = Ax 3 + Bx 3 +Bx 2 + Cx 2 +Cx + Dx+D x 1 = (A+B)x 3 + (B+C)x 2 + (C+D)x+D A + B = 0 A = 2 B + C = 0 B = 2 C + D = 1 C = 2 D = 1 D = 1 x 1 = x 3 (x+1) x+1 x x 2 x
7 x 1 dx x 1 = A + B + C + D x 3 (x+1) x 3 (x+1) x+1 x x 2 x 3 x 1 = x 3 (x+1) x+1 x x 2 x 3 x 1 = 2 ln x+1 2 ln x c x 3 (x+1) x 2x 2 x 1 dx x 3 (x+1) Do we really need x, x 2, and x 3? x 1 = A + B x 3 (x+1) x+1 x 3 x 1 = Ax 3 + B(x+1) x 1 = Ax 3 + Bx+ B A = 0 B = 1 B = 1 contradiction!
8 2x 3 dx (x 1) 2 2x 3 = + #4 (x 1) 2 0 2x 3 dx (x 1) 2 2x 3 = A + B (x 1) 2 x 1 (x 1) 2 2x 3 = A(x 1)+ B = Ax A + B A = 2 B A = 3, B = 1 2x 3 = 2 dx + 1 dx (x 1) 2 x 1 (x 1) 2 = 2 ln x c x
9 8x3 +13x dx (x 2 +2) 2 #5 8x 3 +13x = Ax+B + Cx+D (x 2 +2) 2 x 2 +2 (x 2 +2) 2 3. For numerators, use degree less than degree of the denominator: Denominator is quadratic or power of quadratic, so use a linear form in the numerator. 8x3 +13x dx (x 2 +2) 2 8x 3 +13x = Ax+B + Cx+D = (Ax+B)(x 2 +2) + Cx+D (x 2 +2) 2 x 2 +2 (x 2 +2) 2 (x 2 +2) 2 Ax 3 +2Ax+Bx 2 +2B+Cx+D = Ax 3 +Bx 2 +(2A+C)x+(2B+D) 8x 3 +13x = Ax 3 +Bx 2 +(2A+C)x+(2B+D) A = 8 B 2A +C = 0 = 13 2B +D = 0 A=8 B=0 C= 3 D=0 8x 3 +13x dx = 8x dx + 3x dx (x 2 +2) 2 x 2 +2 (x 2 +2)
10 8x 3 +13x dx = 8x dx + 3x dx (x 2 +2) 2 x 2 +2 (x 2 +2) 2 8x 3 +13x dx = 8 2x 2x (x 2 +2) 2 2 x 2 +2 dx (x 2 +2) 2 dx u=x du 3 u 2 du u 2 4 ln u 3 u 1 + c 2 1 du=2xdx 4 ln x c 2 x x2 +x+1 dx x 3 +x #
11 x2 +x+1 dx x 3 +x To integrate, decompose the fraction x 2 +x+1 = A + Bx+C x(x 2 +1) x x 2 +1 x 2 +x+1 = (A+B)x 2 +Cx+A x(x 2 +1) x(x 2 +1) = A(x 2 +1)+(Bx+C)x = Ax 2 +A+Bx 2 +Cx x(x 2 +1) x(x 2 +1) x 2 +x+1 = (A+B)x 2 +Cx+A A + B = 1 B=0 C = 1 C=1 A = 1 A=1 x 2 +x+1 = A + Bx+C = x(x 2 +1) x x 2 +1 x x 2 +1 x2 +x+1 dx x 3 +x = 1 dx + 1 dx x x 2 +1 x2 +x+1 dx x 3 +x = 1 dx + 1 dx x x 2 +1 = ln x + sec 2 θ dθ sec 2 θ = ln x + dθ = ln x + θ + c tanθ=x sec 2 θdθ=dx secθ = x 2 +1 secθ = x 2 +1 = ln x + arctan x + c
12 x 16x 4 1 dx x dx 16x 4 1 x = x = Ax+B + C + D (4x 2 +1)(4x 2 1) (4x 2 +1)(2x+1)(2x 1) 4x x+1 2x 1 x = (Ax+B)(4x 2 1)+C(4x 2 +1)(2x 1)+D(4x 2 +1)(2x+1) x = 4Ax 3 Ax+4Bx 2 B+C(8x 3 4x 2 +2x 1)+D(8x 3 +4x 2 +2x+1) x = (4A+8C+8D)x 3 +(4B 4C+4D)x 2 +( A+2C+2D)x+( B C+D) 4A +8C+8D = 0 A= 1/2 4B 4C+4D = 0 B=0 A +2C+2D = 1 C=1/8 B C + D = 0 D=1/8 x = x (4x 2 +1)(2x+1)(2x 1) 2(4x 2 +1) 8(2x+1) 8(2x 1) x = x dx + 1 dx + 1 dx (4x 2 +1)(2x+1)(2x 1) 2(4x 2 +1) 8(2x+1) 8(2x 1) 1 ln (4x 2 +1) + 1 ln 2x ln 2x 1 + c Use matrix algebra on calculator
13 x dx 16x 4 1 Different approach to finding constants. x = x = Ax+B + C + D (4x 2 +1)(4x 2 1) (4x 2 +1)(2x+1)(2x 1) 4x x+1 2x 1 x = (Ax+B)(4x 2 1)+C(4x 2 +1)(2x 1)+D(4x 2 +1)(2x+1) x=1/2: 1/2 = (Ax+B)(0)+C(4x 2 +1)(0)+D(2)(2) = 4D; D = 1/8 x= 1/2: 1/2 = (Ax+B)(0)+C(2)( 2)+D(2)(0) = 4C; C = 1/8 x=1: 1 = (A+B)(3)+C(5)(1)+D(5)(3) = 3A + 3B + 5C + 15D x=2: 2 = (2A+B)(15)+C(17)(3)+D(17)(5) = 30A +15B +51C +85D 3A + 3B + 5/8 + 15/8 = 1 3A + 3B = 12/8 = 3/2 30A+15B +51/8 +85/8 = 2 30A+15B = 15 x = x (4x 2 +1)(2x+1)(2x 1) 2(4x 2 +1) 8(2x+1) 8(2x 1) (4x 2 +1)(2x+1)(2x 1) 2(4x 2 +1) 8(2x+1) 8(2x 1) x = x dx + 1 dx + 1 dx 1 ln (4x 2 +1) + 1 ln 2x ln 2x 1 + c B=0, A= 1/2 Matrix Algebra on Calculator to find Constants Line up the coefficients in array. Enter coefficients in matrix A. Enter values in matrix B. Enter the following operation: [A] 1 * [B] using the inverse button, x 1, and enter = 1/8 return to problem
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