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Unformatted text preview: nh−1 2ex + 1 √ 3 +C √ 3 tan θ, 2 √ 2 e2x + ex + 1 2ex + 1 √ sec θ dθ = ln | sec θ + tan θ| + C = ln +√ 3 3 Alternate solution: let ex + 1/2 = + C1 = ln(2 e2x + ex + 1 + 2ex + 1) + C 43. 1 1 dx = 2(x + 1)2 + 5 2 44. 2x + 3 dx, let u = x + 1/2, 4(x + 1/2)2 + 4 1 2u + 2 du = 4u2 + 4 2 = 2 45. 1 √ 1 u + u2 + 1 u2 + 1 du = 2/5(x + 1) + C 1 1 ln(u2 + 1) + tan−1 u + C 4 2 1 1 ln(x2 + x + 5/4) + tan−1 (x + 1/2) + C 4 2 2 1 dx = 4x − x2 1 4 − (x − 2)2 1 1 dx = sin−1 x−2 2 2 = π/6 1 1 4x − x2 dx = 46. 1 1 dx = √ tan−1 (x + 1)2 + 5/2 10 4 − (x − 2)2 dx, let x − 2 = 2 sin θ, 0 0 −π/6 −π/6 2 cos θ dθ = 2θ + sin 2θ 4 −π/2 −π/2 √ 2π 3 − = 3 2 48. u = x sin x, du = (x cos x + sin x) dx; 1 + u2 du = 1 u 2 1 + u2 + 1 1 1 sinh−1 u + C = x sin x 1 + x2 sin2 x + sinh−1 (x sin x) + C 2 2 2 49. u = sin2 x, du = 2 sin x cos x dx; 1 2 1 − u2 du = 1 u 4 1 − u2 + sin−1 u + C = 1 sin2 x 1 − sin4 x + sin−1 (sin2 x) + C 4 Exercise Set 9.5 314 50. u = 3x = ex ln 3 , du = (ln 3)3x dx; 1 ln 3 3 u2 1 1 u u2 − 1 − ln u + − 1 du = 2 ln 3 3 u2 −1 1 √ √ 6 2 − ln(3 + 2 2) = 2 ln 3 EXERCISE SET 9.5 1. B A + (x − 2) (x + 5) 2. 5 A B C =+ + x(x − 3)(x + 3) x x−3 x+3 3. A B C 2x − 3 = + 2+ x2 (x − 1) x x x−1 4. A B C + + x + 2 (x + 2)2 (x + 2)3 5. B A C Dx + E + 2+ 3+ 2 x x x x +1 6. A Bx + C +2 x−1 x +5 7. Cx + D Ax + B +2 x2 + 5 (x + 5)2 8. Bx + C Dx + E A +2 +2 x−2 x +1 (x + 1)2 9. A B 1 1 1 = + ; A = − , B = so (x + 4)(x − 1) x+4 x−1 5 5 − 10. 1 1 dx − x+1 6 1 dx + 3 2x − 1 1 1 1 x+1 1 dx = ln |x + 1| − ln |x + 7| + C = ln +C x+7 6 6 6 x+7 5 1 dx = ln |2x − 1| + 3 ln |x + 4| + C x+4 2 A B 5x − 5 = + ; A = 1, B = 2 so (x − 3)(3x + 1) x − 3 3x + 1 1 dx + 2 x−3 13. 1 1 1 x−1 1 dx = − ln |x + 4| + ln |x − 1| + C = ln +C x−1 5 5 5 x+4 A B 11x + 17 = + ; A = 5, B = 3 so (2x − 1)(x + 4) 2x − 1 x + 4 5 12. 1 1 dx + x+4 5 A B 1 1 1 = + ; A = , B = − so (x + 1)(x + 7) x+1 x+7 6 6 1 6 11. 1 5 2 1 dx = ln |x − 3| + ln |3x + 1| + C 3x + 1 3 A B C 2x2 − 9x − 9 =+ + ; A = 1, B = 2, C = −1 so x(x + 3)(x − 3) x x+3 x−3 1 dx + 2 x 1 dx − x+3 x(x + 3)2 1 dx = ln |x| + 2 ln |x + 3| − ln |x − 3| + C = ln +C x−3 x−3 Note that the symbol C has been recycled; to save space this recycling is usually not mentioned. 315 14. Chapter 9 A B C 1 1 1 =+ + ; A = −1, B = , C = so x(x + 1)(x − 1) x x+1 x−1 2 2 1 1 dx + x 2 − 1 1 dx + x+1 2 1 1 1 dx = − ln |x| + ln |x + 1| + ln |x − 1| + C x−1 2 2 = (x + 1)(x − 1) 1 |x2 − 1| 1 ln +C + C = ln 2 2 x 2 x2 15. 6 x2 + 2 =x−2+ , x+2 x+2 x−2+ 6 x+2 dx = 12 x − 2x + 6 ln |x + 2| + C 2 16. 3 x2 − 4 =x+1− , x−1 x−1 x+1− 3 x−1 dx = 12 x + x − 3 ln |x − 1| + C 2 17. 12x − 22 12x − 22 B A 3x2 − 10 =3+ 2 , + = ; A = 12, B = 2 so − 4x + 4 x − 4x + 4 (x − 2)2 x − 2 (x − 2)2 x2 3dx + 12 18. 1 dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C (x − 2)2 x2 3x − 2 3x − 2 A B =1+ 2 , = + ; A = −1, B = 4 so x2 − 3x + 2 x − 3x + 2 (x − 1)(x − 2) x−1 x−2 1 dx + 4 x−1 dx − 19. 1 dx + 2 x−2 1 dx = x − ln |x − 1| + 4 ln |x − 2| + C x−2 2x2 + x + 1 x5 + 2x2 + 1 = x2 + 1 + , x3 − x x3 − x A B C 2x2 + x + 1 =+ + ; A = −1, B = 1, C = 2 so x(x + 1)(x − 1) x x+1 x−1 1 dx + x (x2 + 1)dx − = 20. 1 dx + 2 x+1 1 dx x−1 1 (x + 1)(x − 1)2 13 x + x − ln |x| + ln |x + 1| + 2 ln |x − 1| + C = x3 + x + ln +C 3 3 x 28x − 1 2x5 − x3 − 1 = 2x2 + 7 + 3 , 3 − 4x x x − 4x A B C 1 57 55 28x − 1 =+ + ;A= ,B=− ,C= so x(x + 2)(x − 2) x x+2 x−2 4 8 8 (2x2 + 7)dx + = 21. 1 4 57 1 dx − x 8 55 1 dx + x+2 8 1 dx x−2 23 57 55 1 x + 7x + ln |x| − ln |x + 2| + ln |x − 2| + C 3 4 8 8 2x2 + 3 B C A + =+ ; A = 3, B = −1, C = 5 so x(x − 1)2 x x − 1 (x − 1)2 3 1 dx − x 1 dx + 5 x−1 1 dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C (x − 1)2 Exercise Set 9.5 22. A B 3x2 − x + 1 C = + 2+ ; A = 0, B = −1, C = 3 so 2 (x − 1) x x x x−1 1 dx + 3 x2 − 23. 316 1 dx = 1/x + 3 ln |x − 1| + C x−1 B C A x2 + x − 16 + + = ; A = −1, B = 2, C = −1 so 2 (x + 1)(x − 3) x + 1 x − 3 (x − 3)2 1 dx + 2 x+1 − 1 dx − x−3 1 dx (x − 3)2 = − ln |x + 1| + 2 ln |x − 3| + 24. 2x2 − 2x − 1 A B C = + 2+ ; A = 3, B = 1, C = −1 so x2 (x − 1) x x x−1 3 25. 1 dx + x 1 dx − x2 1 dx + 4 (x + 2)2 2 4 1 − dx = ln |x + 2| + +C 3 (x + 2) x + 2 (x + 2)2 2x2 + 3x + 3 B A C + = + ; A = 2, B = −1, C = 2 so 3 2 (x + 1) x + 1 (x + 1) (x + 1)3 2 27. 1 1 dx = 3 ln |x| − − ln |x − 1| + C x−1 x B A C x2 + = + ; A = 1, B = −4, C = 4 so 3 2 (x + 2) x + 2 (x + 2) (x + 2)3 1 dx − 4 x+2 26. 1 (x − 3)2 1 + C = ln + +C x−3 |x + 1| x−3 1 dx − x+1 1 dx + 2 (x + 1)2 1 1 1 − dx = 2 ln |x + 1| + +C 3 (x + 1) x + 1 (x + 1)2 2x2 − 1 A Bx + C = +2 ; A = −14/17, B = 12/17, C = 3/17 so 2 + 1) (4x − 1)(x 4x − 1 x +1 7 6 3 2x2 − 1 dx = − ln |4x − 1| + ln(x2 + 1) + tan−1 x + C (4x − 1)(x2 + 1) 34 17 17 28. A Bx + C 1 =+2 ; A = 1, B = −1, C = 0 so + 1) x x +1 x(x2 x3 29. 1 x2 1 1 dx = ln |x| − ln(x2 + 1) +...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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