1. Lesson 7-4 Partial Fractions

2. Fractional Integral Types Type I – Improper: –Degree of numerator ≥ degree of denominator –Start with long division Type II – Proper: –Degree of numerator < degree of denominator –Check common forms or decompose into partial fractions –Repeated terms require special techniques Type III – Variations of Arctan: –U-substitution used a lot Type IV – Variations of Arcsin: –U-substitution used a lot

3. Strategies for Fractional Integrals ExpressionStrategyEvaluates to x³  -------- dx x + 1 ImproperLong DivisionPolynomial + k ln|x-1| + C dx  ----- x³ Proper Check for u n du u n+1 -------- + C n+1 x²  -------- dx x³ + 1 Check for du/u k ln |u| + C 7  -------- dx x² - 1 Partial Fractions Many forms including: k ln|x+1| ± j ln|x-1| + C 5  -------- dx x² + 1 Arc tan k tan -1 u + C 6  ----------- dx  9 – x² Arc sin k sin -1 u + C

4. 7-4 Example 1 x --------- dx = x - 1  1 x-1  x + 0 x - 1 + 1 Long Division 1 = 1 dx + --------- dx x - 1   = x + ln |x – 1| + C

5. 7-4 Example 2 x³ --------- dx = x + 1  x² - x + 1 x+1  x³+ 0x² + 0x + 0 x³ + x² -x² + 0x -x² - x x + 0 x + 1 Long Division = ⅓x³ - ½x² + x - ln |x+1| + C 1 = x² dx - x dx + 1 dx - --------- dx x + 1 

6. Partial Fractions Example 3x – 1 Example: -------------- dx x² - x – 6 I) factor denominator II) rewrite fraction III) multiply through by common denominator IV) solve one factor and substitute V) repeat with remaining factor(s VI) substitute A and B and integrate ∫ x² - x – 6 = (x – 3) (x – 2) 3x – 1 A B ------------- = --------- + ---------- x² – x – 6 (x – 3) (x – 2) 3x – 1 = A(x + 2) + B(x – 3) when x = -2 then B = 7/5 when x = 3 then A = 8/5 8 1 7 1 -- -------- dx + -- --------- dx 5 (x-3) 5 (x + 2) = (8/5) ln|x-3| + (7/5) ln|x+2| + C ∫∫

7. 7-4 Example 4 x + 7 -------------- dx = x² - x - 6  x + 7 ------------------ dx (x - 3)(x + 2)  A B ---------- dx + ---------- dx (x - 3) (x + 2)   B(x - 3) + A(x + 2) = x + 7 B(-5) = -2 + 7 -5B = 5 B = -1 A(5) = 3 + 7 5A = 10 A = 2 2 1 ---------- dx - ---------- dx (x - 3) (x + 2)   2 ln|x-3| - ln|x+2| + C

8. 7-4 Example 5 5x² + 20x + 6 --------------------- dx = x³ + 2x² + x  5x² + 20x + 6 ---------------------- dx x(x + 1)(x + 1)  A B C ------ dx + ---------- dx + ----------- dx (x) (x +1) (x + 1)²   A(x +1)² + B(x)(x + 1) + Cx = 5x² + 20x + 6 A + B = 5 (x²) 2A + B + C = 20 (x) A = 6 (#) 6 + B = 5 B = -1 2(6) + -1 + C = 20 C = 9 6 ln|x| - ln|x+1| - 9/(x+1) + C  6 1 9 ---- dx - ---------- dx + ------------ dx x (x + 1) (x + 1)²   

9. 7-4 Example 6 k k u Variations of Arctan: ---------- du = --- tan -1 (---) + C u² + a² a a  dx -------------- 4x² + 9 dx -------------- (x+1)² + 4 dx ---------------- x² + 4x + 5    u = 2x and a = 3 u = x+1 and a = 2 x² + 4x + 4 + 1 = (x+2)² + 1² u = x+2 and a = 1 1 2x ---- tan -1 (------) + C 6 3 1 x+1 ---- tan -1 (--------) + C 2 2 1 x+2 ---- tan -1 (--------) + C 1 1

10. 7-4 Example 7 k k x Variations of Arcsin: ---------- dx = --- sin -1 (---) + C  a² - x² a a  dx --------------  9 - x² dx --------------  16 – (1+x)² dx ----------------  -4x - x²    u = x and a = 3 u = x+1 and a = 4 4 – 4 – 4x - x² = 2² - (x+2)² u = x+2 and a = 2 1 x ---- sin -1 (------) + C 3 3 1 x+1 ---- sin -1 (--------) + C 4 4 1 x+2 ---- sin -1 (--------) + C 2 2

11. Summary & Homework Summary: –No quotient rule for integration –Compare highest power numerator vs denominator Numerator  Denominator –Use long division to simplify Numerator < Denominator –Find common form –Use partial fractions Homework: –pg 504-505, Day 1: 1, 2, 3, 7 Day 2: 4, 10, 19, 40