CSM_Chapters7 - www.elsolucionario.org Chapter 7 Techniques of Integration 7.1 Integration \u2014 Three Resources 1 Z 5\u22125x dx = \u2212 2 Z \u221a 1 \u221a xe x Z

CSM_Chapters7 - www.elsolucionario.org Chapter 7 Techniques...

This preview shows page 1 out of 128 pages.

Unformatted text preview: Chapter 7 Techniques of Integration 7.1 Integration — Three Resources 1. Z 5−5x dx = − 2. Z √ 1 √ xe x Z 1 5−5x (−5 dx) u = −5x, du = −5 dx 5   Z 1 5−5x 1 1 1 u 5u du = − 5 +C =− + C = − 5x+1 +C =− 5 5 ln 5 5 ln 5 5 ln 5   Z Z 1 −1/2 −1/2 −x1/2 −x1/2 dx = x e dx = −2 e dx − x 2 1 u = −x1/2 , du = − x−1/2 dx 2 Z 1/2 2 = −2 eu du = −2eu + C = −2e−x + C = − √x + C e √   Z Z sin 1 + x 1 −1/2 1/2 √ 3. dx = 2 [sin(1 + x) ] (1 + x) 2 1+x 4. Z 1 u = (1 + x)1/2 , du = (1 + x)−1/2 dx 2 Z √ = 2 sin u du = −2 cos u + C = −2 cos 1 + x + C cos e−x dx = − ex =− 5. Z √ Z Z (cos e−x )(−e−x dx) u = e−x , du = −e−x dx cos u du = − sin u + C = − sin e−x + C Z x 1 dx = − (25 − 4x2 )−1/2 (−8x dx) u = 25 − 4x2 , du = −8x dx 8 25 − 4x2 Z 1 1 1p =− u−1/2 du = − (2u1/2 ) + C = − 25 − 4x2 + C 8 8 4 401 402 CHAPTER 7. TECHNIQUES OF INTEGRATION Z Z 1 1 √ 6. dx = 2 25 − 4x2 Z 1 = 2 Z Z 1 √ dx = 7. x 4x2 − 25 Z = 8. Z 9. Z 10. Z 11. Z 12. Z 13. Z 1 p (2 dx) u = 2x, du = 2 dx 2 5 − (2x)2 1  −1 u  1 1 2x √ du = sin + C = sin−1 +C 2 2 2 5 2 5 5 −u 1 u = 2x, du = 2 dx (2 dx) (2x)2 − 52 u 2x 1 1 1 √ du = sec−1 + C = sec−1 + C 2 2 5 5 5 5 u u −5 2x p Z 1 1 1 p dx = (2 dx) u = 2x, du = 2 dx 2 2 2 25 + 4x 5 + (2x)2 Z  p 1 1 1  √ ln u + u2 + 52 + C = du = 2 2 52 + u2 p 1 = ln 2x + 4x2 + 25 + C 2 Z 1 1 1 u = 2x, du = 2 dx dx = (2 dx) 2 2 25 + 4x 2 5 + (2x)2   Z 1 1 1 1 1 2x −1 u = du = tan +C = tan−1 +C 2 2 2 5 +u 2 5 5 10 5 Z x 1 1 dx = (8x dx) u = 25 + 4x2 , du = 8x dx 2 25 + 4x 8 25 + 4x2 Z 1 1 1 1 = du = (ln |u|) + C = ln |25 + 4x2 | + C 8 u 8 8 Z 1 1 1 dx = (2 dx) u = 2x, du = 2 dx 4x2 − 25 2 (2x)2 − 52   Z u − 5 1 1 1 1 +C = du = ln 2 u2 − 52 2 2(5) u + 5 2x − 5 1 +C = ln 20 2x + 5 Z 1 1 √ p dx = u = 2x, du = 2 dx (2 dx) 2 2 x 4x + 25 2x 5 + (2x)2 √ Z 1 1 5 + 25 + u2 √ = du = − ln +C 5 u u 5 2 + u2 √ 1 5 + 25 + 4x2 = − ln +C 5 2x Z 1 cot 10x dx = cot 10x(10 dx) u = 10x, du = 10 dx 10 Z 1 1 1 = cot u du = (ln | sin u|) + C = ln | sin 10x| + C 10 10 10 √ 403 7.1. INTEGRATION — THREE RESOURCES Z Z 1 14. x csc x dx = (csc2 x2 )(2x dx) u = x2 , du = 2x dx 2 Z 1 1 1 = csc2 u du = (− cot u) + C = − cot x2 + C 2 2 2 Z Z 6 6 u = 3 − 5t, du = −5 dt dx = − 15. (3 − 5t)−2.2 (−5 dt) 2.2 (3 − 5t) 5   Z 6 6 1 −1.2 1 −2.2 u +C =− du = − − u +C = 5 5 1.2 (3 − 5t)1.2 16. Z 2 x 2 p 2 (1 − Z 1 (1 − x3 )5/2 (−3x2 dx) dx = − u = 1 − x3 , du = −3x2 dx 3   Z 1 2 7/2 2p 1 u5/2 du = − u +C =− =− (1 − x3 )7 + C 3 3 7 21 x3 ) 5 Z 1 (sec 3x)(3 dx) u = 3x, du = 3 dx 3 Z 1 1 1 = sec u du = (ln | sec u + tan u|) + C = ln | sec 3x + tan 3x| + C 3 3 3 Z Z 18. 2 csc 2x dx = (csc 2x)(2 dx) u = 2x, du = 2 dx Z = csc u du = ln | csc u − cot u| + C = ln | csc 2x − cot 2x| + C 17. Z sec 3x dx = 19. Z sin−1 x √ dx = 1 − x2 = 20. Z Z Z (sin−1 x) u du = 1 dx = (1 + x2 ) tan−1 x = 21. Z sin x dx = − 1 + cos2 x 22. Z cos(ln 9x) dx = x Z   1 dx 1 − x2 u = sin−1 x, du = √ 1 2 1 u + C = (sin−1 x)2 + C 2 2 Z  Z √ 1 tan−1 x   1 dx 1 + x2 1 dx 1 − x2 1 dx 1 + x2 u = tan−1 x, du = 1 du = ln |u| + C = ln | tan−1 x| + C u 1 (− sin x dx) u = cos x, du = − sin x dx + cos2 x Z 1 =− du = − tan−1 u + C = − tan−1 (cos x) + C 12 + u2 = Z Z 12 [cos(ln 9x)]   1 dx x u = ln 9x, du = 9 cos u du = sin u + C = sin(ln 9x) + C  1 9x  dx = 1 dx x 404 CHAPTER 7. TECHNIQUES OF INTEGRATION Z 1 x3 (sech2 x4 )(4x3 dx) dx = u = x4 , du = 4x3 dx 4 cosh2 x4 Z 1 1 1 = sech2 u du = tanh u + C = tanh x4 + C 4 4 4  Z Z  1 (sinh x dx) u = cosh x, du = sinh x dx 24. tanh x dx = cosh x Z 1 = du = ln |u| + C = ln | cosh x| + C u Z Z 1 u = 2x, du = 2 dx (sec 2x tan 2x)(2 dx) 25. tan 2x sec 2x dx = 2 Z 1 1 1 = sec u tan u du = sec u + C = sec 2x + C 2 2 2 Z Z u = cos x, du = − sin x dx 26. sin x sin(cos x) dx = − [sin(cos x)](− sin x dx) Z = − sin u du = −(− cos u) + C = cos(cos x) + C 23. Z 27. Z sin x csc(cos x) cot(cos x) dx = − Z cos x csc (sin x) dx = 28. 2 = 29. Z Z Z 30. 1 dx = x(ln x)2 = 31. Z Z Z [csc(cos x) cot(cos x)](− sin x dx) u = cos x, du = − sin x Z = − csc u cot u du = −(− csc u) + C = csc(cos x) + C [csc2 (sin x)](cos x dx) u = sin x, du = cos x dx csc2 u du = − cot u + C = − cot(sin x) + C (1 + tan x)2 sec2 x dx u = 1 + tan x, du = sec2 x dx = Z Z (ln x) Z −2 u2 du =  1 dx x  1 1 3 u + C = (1 + tan x)3 + C 3 3 u = ln x, du = u−2 du = −u−1 + C = − 1 dx x 1 +C ln x Z 1 e2x dx = (1 + e2x )−1 (2e2x dx) u = 1 + e2x , du = 2e2x dx 1 + e2x 2 Z 1 1 1 = u−1 du = ln |u| + C = ln(1 + e2x ) + C 2 2 2 7.2. INTEGRATION BY SUBSTITUTION 32. Z 7.2 1. 405 Z 1 (ex dx) u = ex , du = ex dx 12 + (ex )2 Z 1 du = tan−1 u + C = tan−1 ex + C = 12 + u2 ex dx = 1 + e2x Integration by Substitution Z x(x + 1)3 dx = = Z u = x + 1, x = u − 1, dx = du Z 3 (u − 1)u du = (u4 − u3 ) du 1 5 1 4 1 1 u − u + C = (x + 1)5 − (x + 1)4 + C 5 4 5 4 2. Z x2 − 3 dx (x + 1)3 3. Z √ (2x + 1) x − 5 dx 4. Z √ (x2 − 1) 2x + 1 dx 5. Z √ x dx x−1 u = x + 1, x = u − 1, dx = du Z Z 2 Z (u − 1)2 − 3 u − 2u − 2 = du = du = (u−1 − 2u−2 − 2u−3 ) du u3 u3 1 2 + +C = ln |u| + 2u−1 + u−2 + C = ln |x + 1| + x + 1 (x + 1)2 u = x − 5, x = u + 5, dx = du Z 4 22 = (2u + 11)u1/2 du = (2u3/2 + 11u1/2 ) du = u5/2 + u3/2 + C 5 3 4 22 = (x − 5)5/2 + (x − 5)3/2 + C 5 3 Z 1 1 (u − 1), dx = du 2 2    Z  2  1/2 Z  2 1 u − 2u + 1 u (u − 1) 1/2 −1 u du = −1 du = 4 2 4 2   Z u5/2 u3/2 3u1/2 1 7/2 1 1 = − − du = u − u5/2 − u3/2 + C 8 4 8 28 10 4 1 1 1 = (2x + 1)7/2 − (2x + 1)5/2 − (2x + 1)3/2 + C 28 10 4 u = 2x + 1, x = u = x − 1, x = u + 1, dx = du Z Z u+1 2 = du = (u1/2 + u−1/2 ) du = u3/2 + 2u1/2 + C 1/2 3 u 2 3/2 1/2 = (x − 1) + 2(x − 1) + C 3 406 CHAPTER 7. TECHNIQUES OF INTEGRATION 6. Z √ 7. Z x+3 dx (3x − 4)3/2 8. Z √ (x2 + x) 3 x + 7 dx x2 dx x+2 u = x + 2, x = u − 2, dx = du Z Z (u − 2)2 = du = (u3/2 − 4u1/2 + 4u−1/2 ) du u1/2 2 8 2 8 = u5/2 − u3/2 + 8u1/2 + C = (x + 2)5/2 − (x + 2)3/2 + 8(x + 2)1/2 + C 5 3 5 3 1 1 u = 3x − 4, x = (u + 4), dx = du 3 3   Z Z u/3 + 13/3 1 1 −1/2 13 −3/2 2 26 = du = u + u = u1/2 − u−1/2 + C 3/2 3 9 9 9 9 u 2 26 1/2 −1/2 = (3x − 4) − (3x − 4) +C 9 9 = Z Z u = x + 7, x = u − 7, dx = du Z [(u − 7)2 + (u − 7)]u1/3 du = (u2 − 13u + 42)u1/3 du 3 10/3 39 7/3 63 4/3 u − u + u +C 10 7 2 3 39 63 10/3 7/3 4/3 = (x + 7) − (x + 7) + (x + 7) + C 10 7 2 = √ 9. Z x dx x+1 10. Z √ √ x, x = u2 , dx = 2u du  Z Z  Z 2 2u2 u (2u du) = du = 2− 2 du = u2 + 1 u2 + 1 u +1 √ √ = 2u − 2 tan−1 u + C = 2 x − 2 tan−1 x + C t dt t+1 Z √ t−3 √ 11. dt t+1 u= (u7/3 − 13u4/3 + 42u1/3 ) du = u= √ u= √ t + 1, t = (u − 1)2 , dt = 2(u − 1) du Z Z 2(u − 1)3 (u − 1)2 [2(u − 1) du] = du = u u   Z 2 2 = 2u2 − 6u + 6 − du = u3 − 3u2 + 6u − 2 ln |u| + C u 3 √ √ √ 2 √ = ( t + 1)3 − 3( t + 1)2 + 6( t + 1) − 2 ln( t + 1) + C 3 t + 1, t = (u − 1)2 , dt = 2(u − 1) du  Z Z  u−4 8 = [2(u − 1) du] = 2u − 10 + du = u2 − 10u + 8 ln |u| + C u u √ √ √ = ( t + 1)2 − 10( t + 1) + 8 ln( t + 1) + C 7.2. INTEGRATION BY SUBSTITUTION 12. Z √ r+3 dr r+3 u= √ r, r = u2 , dr = 2u du  Z Z Z  6 u+3 2u2 + 6u 6u = (2u du) = du = − du 2+ 2 u2 + 3 u2 + 3 u + 3 u2 + 3 u 6 = 2u + 3 ln(u2 + 3) − √ tan−1 √ + C 3 3 r √ √ r = 2 r + 3 ln(r + 3) − 2 3 tan−1 +C 3 Z Z x3 x2 √ (x dx) 13. dx = u = x2 + 1, du = 2x dx 3 2 2 (x + 1)1/3 x +1   Z Z 3 5/3 3 2/3 u−1 1 1 (u2/3 − u−1/3 ) du = du = u − u +C = 1/3 2 2 10 4 u 3 3 2 (x + 1)5/3 − (x2 + 1)2/3 + C = 10 4 Z Z x5 x4 √ 14. dx = (x dx) u = x2 + 4, du = 2x dx 5 (x2 + 4)1/5 x2 + 4   Z Z (u − 4)2 1 1 = du = (u9/5 − 8u4/5 + 16u−1/5 ) du 2 2 u1/5 5 14/5 20 9/5 u − u + 10u4/5 + C = 28 9 5 2 20 14/5 = (x + 4) − (x2 + 4)9/5 + 10(x2 + 4)4/5 + C 28 9 Z x2 15. dx u = x − 1, du = dx (x − 1)4 Z Z (u + 1)2 1 = du = (u−2 + 2u−3 + u−4 ) du = −u−1 − u−2 − u−3 + C u4 3 1 1 1 =− − − +C 2 x − 1 (x − 1) 3(x − 1)3 Z 2x + 1 dx 16. u = x + 7, du = dx (x + 7)2 Z Z 2(u − 7) + 1 = du = (2u−1 − 13u−2 ) du = 2 ln |u| + 13u−1 + C u2 13 +C = 2 ln |x + 7| + x+7 Z √ √ 2u 17. ex − 1 dx u = ex − 1, x = ln(u2 + 1), dx = 2 du u +1   Z  Z  2u 2u = u du = 2− 2 du = 2u − 2 tan−1 u + C u2 + 1 u +1 √ √ = 2 ex − 1 − 2 tan−1 ex − 1 + C 407 408 18. CHAPTER 7. TECHNIQUES OF INTEGRATION Z √ 1 ex − 1 u= dx Z = 19. Z q 20. Z 1− √ v dv 2u u2 + 1  du = Z 2u du u2 + 1 √ 2 du = 2 tan−1 u + C = 2 tan−1 ex − 1 + C u2 + 1 √ v, v = (1 − u)2 , dv = −2(1 − u) du Z Z 4 4 = u1/2 [−2(1 − u) du] = (2u3/2 − 2u1/2 ) du = u5/2 − u3/2 + C 5 3 √ √ 4 4 = (1 − v)5/2 − (1 − v)3/2 + C 5 3 √ w p √ dw 1− w Z u=1− √ w, w = (1 − u)2 , dw = −2(1 − u) du Z 1−u = [−2(1 − u) du] = −2 (u−1/2 − 2u1/2 + u3/2 ) du u1/2   4 3/2 2 5/2 1/2 = −2 2u − u + u + C 3 5 √ √ √ 1/2 8 4 = −4(1 − w) + (1 − w)3/2 − (1 − w)5/2 + C1 3 5 u=1+ = Z √q √ t 1 + t t dt 22. 23. ex − 1, x = ln(u2 + 1), dx = u=1− √ Z p 1+ t √ 21. dt t Z  1 u √ Z √ √ 1 t, du = √ dt 2 t u(2 du) = √ 4 3/2 4 u + C = (1 + t)3/2 + C 3 3 √ 3 u = 1 + t t = 1 + t3/2 , du = t1/2 dt 2 Z √ 2√ 4 4 = u du = u3/2 + C = (1 + t t)3/2 + C 3 9 9 Z Z 2(x + 1) + 5 2(x + 1) 5 dx = dx + dx (x + 1)2 + 4 (x + 1)2 + 4 (x + 1)2 + 4 5 x+1 = ln[(x + 1)2 + 4] + tan−1 +C 2 2 5 x+1 = ln(x2 + 2x + 5) + tan−1 +C 2 2 2x + 7 dx = x2 + 2x + 5 Z 7.2. INTEGRATION BY SUBSTITUTION 24. Z Z 6x − 1 3 2(x + 1/2) − 4/3 dx = dx 2 4x + 4x + 10 4 (x + 1/2)2 + 9/4 Z Z 3 2(x + 1/2) 1 = dx − dx 4 (x + 1/2)2 + 9/4 (x + 1/2)2 + 9/4 2 x + 1/2 3 +C = ln[(x + 1/2)2 + 9/4] − tan−1 4 3 3/2 3 4x2 + 4x + 10 2 2x + 1 = ln − tan−1 +C 4 4 3 3 3 2 2x + 1 = ln(4x2 + 4x + 10) − tan−1 + C1 4 3 3 25. Z 2x + 5 √ dx = 16 − 16x − x2 Z 2(x + 3) − 1 p dx 25 − (x + 3)2 Z Z 2(x + 3) 1 p p = dx − dx 2 25 − (x + 3) 25 − (x + 3)2 = Z 4x − 3 dx = 11 + 10x − x2 Z u = 25 − (x + 3)2 , du = −2(x + 3) dx √ −1 x+3 x+3 √ du − sin−1 + C = −2 u − sin−1 +C 5 5 u p x+3 = −2 16 − 6x − x2 − sin−1 +C 5 26. Z √ 4(x − 5) + 17 p dx 36 − (x − 5)2 Z Z 17 4(x − 5) p p = dx + dx 2 36 − (x − 5) 36 − (x − 5)2 Z u = 36 − (x − 5)2 , du = −2(x − 5) dx √ −2 x−5 x−5 √ du + 17 sin−1 + C = −4 u + 17 sin−1 +C 6 6 u p x−5 = −4 11 + 10x − x2 + 17 sin−1 +C 6 = 27. Z √ 1 √ dx x− 3x u = x1/6 , x = u6 , dx = 6u5 du  Z Z  6u5 1 2 = du = 6 u +u+1+ du u3 − u2 u−1 = 2u3 + 3u2 + 6u + 6 ln |u − 1| + C √ √ √ √ = 2 x + 3 3 x + 6 6 x + 6 ln | 6 x − 1| + C 409 410 CHAPTER 7. TECHNIQUES OF INTEGRATION 28. Z 29. Z √ 3 √ 6 x dx x+1 u = x1/6 , x = u6 , dx = 6u5 du  Z Z Z  u 6u6 6 5 4 2 = (6u du) = du = du 6u − 6u + 6 − 2 u2 + 1 u2 + 1 u +1 6 = u5 − 2u3 + 6u − 6 tan−1 u + C 5 6 = x5/6 − 2x1/2 + 6x1/6 − 6 tan−1 x1/6 + C 5 1 √ x 5x + 4 dx 0 √ x 3 x + 1 dx 1 1 (u − 4), dx = du 5 5   Z 9 Z 9 1 1 1 = (u3/2 − 4u1/2 ) du (u − 4)u1/2 du = 5 5 25 4 4  9     1 486 64 64 506 1 2 5/2 8 3/2 = u − u − 72 − − = = 25 5 3 25 5 5 3 375 4 u = 5x + 4, x = 0 30. Z u = x + 1, du = dx −1 = Z 0 = 31. Z 16 1 32. Z 9 4 33. Z 2 9 1 √ dx 10 + x √ x−1 √ dx x+1 5x − 6 √ dx 3 x−1  1 (u − 1)u1/3 du = 3 7/3 3 4/3 u − u 7 4 u = 10 + 1 0 Z 1 0 = (u4/3 − u1/3 ) du 3 3 9 − =− 7 4 28 √ x, x = (u − 10)2 , dx = 2(u − 10) du  Z 14  Z 14 20 2(u − 10) 14 du = du = (2u − 20 ln |u|)]11 2− = u u 11 11 14 = (28 − 20 ln 14) − (22 − 20 ln 11) = 6 − 20 ln 11 u= √ u= √ 3 x + 1, x = (u − 1)2 , dx = 2(u − 1) du  Z 4 Z 4 4 u−2 4 = [2(u − 1) du] = du = u2 − 6u + 4 ln |u| 3 2u − 6 + u u 3 3 4 = (−8 + 4 ln 4) − (−9 + 4 ln 3) = 1 + 4 ln 3 Z x − 1, x = u3 + 1, dx = 3u2 du 5(u3 + 1) − 6 = (3u2 du) = u 1 3 177 = 90 − = 2 2 2 Z 1 2 (15u − 3u) du = 4  3 3u − u2 2 5 2 1 411 7.2. INTEGRATION BY SUBSTITUTION 34. Z 35. Z 0 √ − 3 1 0 √ 2x3 dx = x2 + 1 (1 − Z 0 x2 (2x dx) u = x2 + 1, du = 2x dx x2 + 1 − 3  1 Z 1 Z 1 u−1 2 3/2 1/2 −1/2 1/2 (u − u ) du = = du = u − 2u 1/2 3 4 u 4 4 4 4 8 =− − =− 3 3 3 √ √ u=1− x)50 dx = √ Z √ x, x = (1 − u)2 , dx = −2(1 − u) du Z 0 u [−2(1 − u) du] =   1 1 = =0− − 1326 1326 50 1 36. Z 4 (1 + 0 1 √ x)3 u=1+ dx Z 8 x1/3 1 38. Z 64 1 1 dx + x2/3 x1/3 dx x2/3 + 2 Z 0 1 1 − 2u ) du = 50  1 52 2 u − u51 26 51 0 1 x, x = (u − 12 ), dx = 2(1 − u) du 3 u = x1/3 , x = u3 , dx = 3u2 du  Z 2 Z 2 3 3u2 2 du = du = (3u − 3 ln |1 + u|)]1 = 3 − 2 u + u 1 + u 1 1 2 = (6 − 3 ln 3) − (3 − 3 ln 2) = 3 + 3 ln 3 u = x1/3 , x = u3 , dx = 3u2 du    4 Z 4 u 6u 3 2 2 2 (3u du) = 3u − du = u − 3 ln(u + 2) 2 2 u +2 2 1 u +2 1 1   45 3 − 3 ln 3 = − 3 ln 6 = (24 − 3 ln 18) − 2 2 = 39. (2u 51 Z 3 3 2(u − 1) du = (2u−2 − 2u−3 ) du = (−2u−1 + u−2 ) 1 3 u 1 1  3 1 2 5 4 = − = − − (−1) = u2 u 1 9 9 = 37. Z √ 0 Z x2 (1 − x)5 dx = 4 Z 1 =  u = 1 − x, du = −dx 0 (1 − u)2 u5 (−du) = 1 6 2 7 1 8 u − u + u 6 7 8 Z 1 0 1 0 = (u5 − 2u6 + u7 ) du 1 2 1 1 − + = 6 7 8 168 412 40. CHAPTER 7. TECHNIQUES OF INTEGRATION Z 6 0 41. Z 2x + 5 √ dx 2x + 4 x2 1 dt t 1 Z x 1 42. Z x 1 1 dt t Z x 1 43. A = 16 √ t, t = u2 , dt = 2u du Z x Z x 1 1 1 (2u du) = 2 du = 2 dt 2 u 1 u 1 t 1 u, dt = √ du 2 u   Z Z 1 1 x1 1 x1 1 √ √ du = du = dt 2 1 u 2 1 t u 2 u u = t2 , t = = Z    16 Z u+1 1 1 16 1/2 1 2 3/2 −1/2 1/2 √ = (u + u du = ) du = u + 2u 2 2 4 2 3 u 4 4     1 128 16 62 = +8 − +4 = 2 3 3 3 Z u= = √ u = 2x + 4, du = 2 dx 1 1 √ u = x1/3 , x = u3 , dx = 3u2 du  Z 1 Z 1 3u2 3 = du = 3u − 3 + du u+1 0 u+1 0  1   3 3 2 = − + 3 ln 2 − 0 ≈ 0.5794 = u − 3u + 3 ln |u + 1| 2 2 0 x1/3 0 44. A = − =− =− Z 0 +1 dx √ x3 x + 1 dx + −1 Z 1 1 1 √ x3 x + 1 dx 1 u = x + 1, du = dx 1 0 (u − 1)3 u1/2 du + 0 Z Z 1 (u 7/2 0 − 3u 5/2 Z 2 1 + 3u 3/2 (u − 1)3 u1/2 du −u 1/2 ) du + Z 1 -1 2 (u7/2 − 3u5/2 + 3u3/2 − u1/2 ) du 1  2 2 9/2 6 7/2 6 5/2 2 3/2 2 9/2 6 7/2 6 5/2 2 3/2 u − u + u − u + u − u + u − u 9 7 5 3 9 7 5 3 0 1 √      √ 52 32 64 + 52 2 32 −0 + 2− − = ≈ 0.4366 =− − 315 315 315 315 =−  45. V = 2π Z 4 0 = 2π Z 1 3 1 dx x+1 u= √ x + 1, x = (u − 1)2 , dx = 2(u − 1) du  Z 3 (u − 1)2 2 2 [2(u − 1) du] = 2π 2u − 6u + 6 − du u u 1 x√ 1 413 7.3. INTEGRATION BY PARTS  3   2 3 11 2 = 2π u − 3u + 6u − 2 ln |u| = 2π (9 − 2 ln 3) − 1 3 3 1   16 − 2 ln 3 ≈ 19.7048 = 2π 3 Z 4 √ 1 √ dx u = x + 1, x = (u − 1)2 , dx = 2(u − 1) du 46. V = π 2 0 ( x + 1)   3 Z 3 Z 3 2(u − 1) 2 2 2 =π du = π du = π 2 ln |u| + − 1 u2 u u2 u 1 1 1      4 2 − 2 = π 2 ln 3 − ≈ 2.7140 = π 2 ln 3 + 3 3 47. y = x1/4 Z 9q Z L= 1 + (x1/4 )2 dx = 0 u=1+ √ 0 9 q 1+ √ x dx x, x = (u − 1)2 , dx = 2(u − 1) du 4  Z 4 Z 4 √ 2 5/2 2 3/2 u − u = u[2(u − 1) du] = 2 (u3/2 − u1/2 ) du = 2 5 3 1 1 1     64 16 2 2 232 =2 − − − = 5 3 5 3 15 Z w2 1 48. T = u = w1/3 , w = u3 , dw = 3u2 du dw 2/3 − Bw Aw w1 Z w21/3 Z w21/3 3 3u2 du = du = 2 − Bu3 1/3 1/3 Au A − Bu w1 w1 w21/3 1/3 3 3 A − Bw1 = − ln |A − Bu| = ln . 1/3 B B A − Bw1/3 w1 2  3 A 1/3 Assuming that A − Bw2 > 0 we see that w2 < . B 7.3 Integration by Parts 1. Z √ x x + 3 dx 2. Z √ x dx 2x − 5 2 u = x, du = dx; dv = (x + 3)1/2 dx, v = (x + 3)3/2 3 Z 2 2 2 4 = x(x + 3)3/2 − (x + 3)3/2 dx = x(x + 3)3/2 − (x + 3)5/2 + C 3 3 3 15 u = x, du = dx; dv = (2x − 5)−1/2 dx, v = (2x − 5)1/2 Z 1 = x(2x − 5)1/2 − (2x − 5)1/2 dx = x(2x − 5)1/2 − (2x − 5)3/2 + C 3 4 4 414 CHAPTER 7. TECHNIQUES OF INTEGRATION 3. Z ln 4x dx 4. Z ln(x + 1) dx 1 u = ln 4x, du = dx; dv = dx, v = x x Z = x ln 4x − dx = x ln 4x − x + C u = ln(x + 1), du = = x ln(x + 1) − Z 1 dx; x+1 dv = dx, v = x x dx = x ln(x + 1) − x+1 = x ln(x + 1) − x + ln(x + 1) + C 5. Z u = ln 2x, du = x ln 2x dx = 6. Z x1/2 ln x dx 7. Z ln x dx x2 8. Z ln x √ dx x3 1 2 x ln 2x − 2 Z x2 1 1 dx = x2 ln 2x − 2x 2 2 Z 1− 1 x+1  dx 1 2 x 2 x dx = 1 dx; dv = x1/2 dx, v = x   Z 2 2 2 x3/2 dx = x3/2 ln x − = x3/2 ln x − 3 3 x 3 2 3/2 4 3/2 = x ln x − x + C 3 9 1 2 1 x ln 2x − x2 + C 2 4 2 3/2 x 3 Z 2 x1/2 dx 3 1 1 1 u = ln x, du = dx; dv = 2 dx, v = − x x x  Z  1 1 1 1 − 2 dx = − ln x − + C = − ln x − x x x x u = ln x, du = −1/2 ln x − Z = −2x−1/2 ln x + 2 9. dv = x dx, v = u = ln x, du = = −2x Z 1 dx; x Z  (ln t)2 dt 1 dx; x Z −2x−1/2 dx x x−3/2 dx = −2x−1/2 ln x − 4x−1/2 + C u = (ln t)2 , du = Z dv = x−3/2 dx, v = −2x−1/2 2 ln t dt; t dv = dt, v = t 2 u = 2 ln t, du = dt; dv = dt, v = t t   Z = t(ln t)2 − 2t ln t − 2 dt = t(ln t)2 − 2t ln t + 2t + C = t(ln t) − 2 2 ln t dt 415 7.3. INTEGRATION BY PARTS 10. Z u = (ln t)2 , du = (t ln t)2 dt 11. Z dv = t2 dt, v = 1 3 t 3 Z 1 2 2 2 2 u = ln t, du = dt; dv = t2 dt, v = t3 t ln t dt 3 t 3 9   Z 1 2 3 1 2 2 2 2 = t3 (ln t)2 − t ln t − t dt = t3 (ln t)2 − t3 ln t + t3 + C 3 9 9 3 9 27 = 1 3 t (ln t)2 − 3 2 ln t dt; t sin−1 x dx u = sin−1 x, du = √ Z 1 dx; 1 − x2 dv = dx, v = x x dx u = 1 − x2 , du = −2x dx 1 − x2   Z √ 1 1 √ = x sin−1 x − − du = x sin−1 x + u + C 2 u p −1 = x sin x + 1 − x2 + C = x sin−1 x − 12. Z x2 tan−1 x dx 13. Z xe3x dx 14. Z x2 e5x dx √ 1 1 dx; dv = x2 dx, v = x3 1 + x2 3    Z Z  3 1 3 x 1 x 1 1 3 −1 −1 = x tan x − x− dx = x tan x − dx 3 3 1 + x2 3 3 1 + x2 1 1 1 = x3 tan−1 x − x2 + ln(1 + x2 ) + C 3 6 6 u = tan−1 x, du = 1 u = x, du = dx; dv = e3x dx, v = e3x 3 Z 1 1 1 1 3x e dx = xe3x − e3x + C = xe3x − 3 3 3 9 u = x2 , du = 2x dx; Z 1 5x e 5 2 1 2 x, du = dx; dv = e5x dx, v = e5x 5 5 5   Z 2 5x 2 5x 1 2 2 5x 1 = x2 e5x − xe − e dx = x2 e5x − xe5x + e +C 5 25 25 5 25 125 = 1 2 5x x e − 5 2 5x xe dx 5 dv = e5x dx, v = u= + − 416 CHAPTER 7. TECHNIQUES OF +INTEGRATION 15. Z 16. Z 17. Z 18. Z 6x 6 1 1 − e−4x 6 − e−4x e 6x 4 16 1 3 3 3 1 1 1 3 −4x 3 −4x 2 −4x −4x −4x 3 2 −4x −4x −4x −4 15. x 3x 6x 6 x e dx = − x e − x e − xe − e +C + e − e 5 e − e 16. 5x 20x3 64160x 41x 1641 4 16 32 128 −4x −15. x3 e − e− 3x5e2 −4x 6x4 − 6e−4x x e 3 + 420x 16 2 16. x 1 – 5x 60x 120x 64 11 1 1 − 15. x3 e−4x3x2 x−5 6xe−4x 46 e−4x − 2 e−4x 3 16.e x 4 5x 60x 1620x 64 120x 25 + 1 1 1 −4x 1 x −4x −4x −4x 2 − + 15. x3 16. 5 4 3 2 e e 3x − e −120x e 6x e 620x x 5x 120 256 e 4 16 60x 64 + 1 −4x 1 −4x 1 −4x − x 1 −4x −4x e4 e e 3 e 60x2 − 120x 15. x3 3xe216. x56xe −5x 20x16 120 46 64 256 + 1 1 1 1 3 2 −4x −4x −4x −4x −4x 15. 6e 3 − 4 e x 16. x3x −5 ex 5x6x −2 e e – 60x 64+1 120x 4 1 −4x 1620x 256120 1 −4x 1 −4x −4x −4x e − e e − e e x − 2 420x3 + 60x 16 2 256 15. x3 3x16. x56x 5x6e4 120x 64 120 1 x−4x5 1 1 −4x 2 − 1 60x 5x4e−4x 20x3− 120x ee x e−4x − 16. e−4x 15.+e x3 1203x2 6x 6 4 x 16 64 256 1 −4x 1 e − 3 2 −4x 15. 3x 6xe 6 − e 2 e 16. x5 5x4 20x3 60x+ 120xx 120 4 16 1 −4x3 1 2−4x 1 − − e−4x − 15. e+ x5 e 3xe4 6x − 6 ex 5x 20x3 64 60 416. x 16 1 −4x 1 − + 15. x3 3x52 6x4 6 e−4x − e e x 3 2 e 16. x 5x 20x 60x 4 120x 16 1 + 1 1 −4x − 5 x 5 x 4 x 3 x 2 x x x e−4x ex − e−4x e 3 6 15.+ x35e−4x 3x24 − 6x x e dx = x e − 5x e + 20x e − 60x e + 120xe − 120e + C 4 16. 16 x 5x 1 6420x 160 − −4x + 15. x3 3x25 6x 4 6 ex 2 − e−4x e 3 e60x 16. x 1 5x 120x 4 16 120x 1120−4x − –e−4x −4x 3 − 13x 2 e−4x − e−4x e e 15. x 6x 6 + x 3 e 4 16 16. x5 645x4 1 20x 256 160 − e−4x 2 2 13 2 2 − e−4x e x 2 x3 ex dx u = x2 , du = 2x dx; dv = xex +dx, v15. = x ex16.3xx5 1 6x5x4 6 120x3 e 60x 1 120x 1 120 4 16 − +− 15. 3 2 e−4x6x e2−4x −x e−4x e−4x e−4x x 3x 6 + 5 4 3 e 4 Z 16 64 256 16. x 5x 20x 60 1 1 − e−4x 2 2 1 1 23 1 2 x2 − e−4x e x 2 e = x2 ex − xex dx = x2 e+ − 15.exx + 16. C3xx5 6x5x4 6 20x3 60x 120x 120 16 1 −4x 1 −4x 1 −4x – 15.+1x3e−4x 3x2 4 6x − 2 2 2 e−4x x − e e − e 6 e 4 16.256 x5 5x4 20x3 60 16 64 1 1 − −4x 15. x3 3x25 6x 4 6 ex − e−4x e 3 2 e 16. x 5x 20x 60x 120x 120 4 16 11 −4x3 1 −4x 1+ −4x 3 1 −4x 3 2 −4x 2 2x 3 2 2x 5 2x3 15. x 3x 6x 6 e − e e − e e x u = x , du = 3x dx; dv = x e e 4dx, v = 16 e x e dx 5x4 1 20x3 64 256 16. x5 160 6 e−4x − e−4x e x 5 4 3 2 Z e 16. x 5x 20x 60x 120x 120 4 16 – + − = 19. 20. 21. Z Z Z −+ − 15. x3 t cos 8t dt 1 3 2x3 x e − 6 15. x3 3x2 3x2 + 3 3 1 2 2x3 1 1 x e dx = exx 3 e2x − e2x + C 2 6 12 16. x5 5x4 20x3 ex 1 u = t, du = dt; dv = cos 8t dt, v = sin 8t 8 Z 1 1 1 1 = t sin 8t − sin 8t dt = t sin 8t + cos 8t + C 8 8 8 64 1 1 1 1 1 x sinh x dx x2 sin x dx u = x, du = dx; dv = sinh x dx, v = cosh x Z = x cosh x − cosh x dx = x cosh x − sinh x + C 1 1 1 u = x2 , du = 2x dx; dv = sin x dx, v = − cos x Z = −x2 cos x + 2x cos x dx u = 2x, du = 2 dx; dv = cos x dx, v = sin x Z = −x2 cos x + 2x sin x − 2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C 1 1 1 1 1 1 60 7.3. INTEGRATION BY PARTS 22. Z x2 cos 23. Z x3 cos 3x dx x dx 2 x x u = x2 , du = 2x dx; dv = cos dx, v = 2 sin 2 2 Z x x = 2x2 sin − 4 x sin dx 2 2 x x u = x, du = dx; dv = sin dx, v = −2 cos 2 2   Z x x x = 2x2 sin − 4 −2x cos + 2 cos dx 2 2 2   x x x = 2x2 sin − 4 −2x cos + 4 sin +C 2 2 2 x x x = 2x2...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture