Bài tập ôn thi Toán: Nguyên hàm tích phân

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Đây là bài tập nguyên hàm tích phân gửi đến các bạn học sinh tham khảo để củng cố kiến thức toán 12.

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Nội dung Text: Bài tập ôn thi Toán: Nguyên hàm tích phân

ch−¬ng 4. nguyªn hμm vμ tÝch ph©n 1. Dïng b¶ng tÝch ph©n c¬ b¶n vµ c¸c tÝnh chÊt cña tÝch ph©n, tÝnh: (1 + x) 2 2) ∫ 1) ∫ (a + b ) dx dx x x2 x(1 + x 2 ) dx 4) ∫ 3) ∫ x . x + 2 .dx 25 3 xcos (1 + lnx) 2 1 + tg 2 x sinxdx 6) ∫ ∫ dx 5) 1 + tgx 1 + 2cosx e x dx x + (arccos3x) 2 8) ∫ 2x ∫ dx 7) e +4 1 − 9x 2 dx dx 10) ∫ ∫ 1 + sinx 9) x 1+ x2 (1 − x) 2 dx 1+ x − x2 12) ∫ ∫ dx 11) xx (1 − x 2 ) 3 Gi¶i. 1) ∫ (a + b ) dx = ∫ [(a ) + 2(ab) + ( b ) ]dx x x2 2x x 2x 2(ab) x a 2x b 2x = + + +C lnb 2 lna 2 ln(ab) (1 + x) 2 ⎛1 2⎞ 2) ∫ dx = ∫ ⎜ + 2⎟ dx = ln x + 2arctgx + C. x(1 + x ) ⎝ x 1+ x ⎠ 2 6 15 3 15 3 3) ∫ x . x + 2 .dx = ∫ . x + 2 .d(x + 2) = . .( x + 2) 5 + C 3 25 3 3 36 55 3 ( x + 2) 6 + C . = 18 d(1 + lnx) dx ∫ xcos 2 (1 + lnx) = ∫ cos 4) = tg(1 + lnx) + C. (1 + lnx) 2
1 d(1 + 2cosx) sinxdx ∫ = ∫− . = − 1 + 2 cos x + C . 5) 1 + 2cosx 2 1 + 2cosx 1 + tg 2 x d(1 + tgx) ∫ 6) ∫ = 2 1 + tgx + C . dx = 1 + tgx 1 + tgx x + (arccos3x) 2 7) ∫ dx 1 − 9x 2 1 1 1 − = − ∫ (1 − 9 x ) 2 d (1 − 9 x ) − ∫ (arccos 3 x) d (arccos 3 x) 2 2 2 3 18 1 1 =− 1 − 9 x 2 − (arccos 3 x) 2 + C . 9 9 d(e x ) e x dx 1 ex ∫ (e x ) 2 + 2 2 = 2 arctg 2 + C. 8) ∫ 2x = e +4 dx dx dx ∫ 1 + sinx = ∫ ∫ 9) = ⎛π ⎞ ⎛π x⎞ 1 + cos ⎜ − x ⎟ 2cos 2 ⎜ − ⎟ ⎝2 ⎠ ⎝4 2⎠ ⎛π x⎞ − d⎜ − ⎟ ⎛π x⎞ ⎝4 2⎠ =∫ = tg ⎜ − ⎟ + C. ⎛π x⎞ ⎝4 2⎠ cos 2 ⎜ − ⎟ ⎝4 2⎠ ) ( 1⎛ ⎞ dx xdx 1 1 ∫x ∫x ∫ ⎜ x 2 +1 −1 x 2 +1 +1⎟ d x +1 ⎜ − 2 ⎟ 10) = = 2⎝ 1+ x 1+ x 2 2 2 ⎠ x 2 +1 −1 1 +C. = ln 2 x2 +1 +1 Chó ý ta còng cã thÓ tÝnh:
dx ∫x = 1+ x2 ⎛1⎞ dx d⎜ ⎟ 2 ⎝ x ⎠ = − ln 1 + 1 + ⎛ 1 ⎞ + C = ln x x2 ∫ x2 +1 = −∫ +C ⎜⎟ ⎝x⎠ x 1+ 1+ x 2 2 ⎛ 1⎞ 1+ ⎜ ⎟ x ⎝x⎠ 1+ x − x2 1− x2 x 11) ∫ dx = ∫ dx + ∫ dx (1 − x ) (1 − x ) (1 − x 2 ) 3 23 23 1 d(1 − x 2 ) dx −∫ ∫ = 2 (1 − x 2 ) 3 1− x2 1 +C. = arcsinx + 1− x 2 ⎛ −3 ⎞ 2x 2 − 12x − 6 (1 − x) 2 dx 1 1 − 12) ∫ = ∫⎜ ⎜ x 2 − 2x 2 + x 2 ⎟dx = +C ⎟ xx 3x ⎝ ⎠ 2. Dïng ph−¬ng ph¸p ®æi biÕn tÝnh c¸c tÝch ph©n sau: dx 2) ∫ 1) ∫ x (1 − 2x ) dx 3 43 (x + 1) x xdx sin4xdx 3) ∫ 4) ∫ cos 2 2x + 4 x 4 −1 3x + 5 ∫ 6) ∫ tg x.dx 3 dx 5) x x2 − x 8) ∫ 7) ∫ x a − x .dx dx 3 2 (x − 2) 3 x − 1 dx dx ∫ ∫ x + 1. x 2 9) 10) x − x2 x +1 dx 11) ∫ ∫ dx 12) x(1 + xe x ) 1+ ex x2 −1 x2 −1 13) ∫ 4 14) ∫ 2 dx dx x +1 (x + 5x + 1)(x 2 − 3x + 1)
Gi¶i. 1 1) ∫ x (1 − 2x ) dx . §Æt t = 1 - 2x4, th× dt = - 8x3dx ⇒ x3dx = - dt. VËy: 3 43 8 13 1 1 ∫ x (1 − 2x ) dx = - 8 ∫ t dt = − 32 t4 + C = - 32 (1 - 2x4)4 + C. 3 43 dx ∫ (x + 1) x . §Æt x = t, th× dx = 2tdt vµ 2) 2tdt dt ∫ (t 2 + 1)t = 2∫ 2 = 2arctgt + C = 2arctg x + C. t +1 xdx ∫ . §Æt t = x2, th× dt = 2xdx. 3) x −1 4 VËy xdx 1 dt 1 1 ∫ ∫ ln t + t − 1 +C = ln x + x − 1 + C = 2 2 4 = 2 2 2 x 4 −1 t 2 −1 sin4xdx ∫ cos 2 2x + 4 . §Æt t = cos2x th× dt = - 2sin2xdx vµ ta cã: 4) sin4xdx 2sin2xcos2 xdx ∫ cos 2 2x + 4 =∫ cos 2 2x + 2 2 1 d(t 2 + 4) 1 tdt =− ∫ 2 −∫ 2 = − ln(t2 + 4) + C = 2 t +2 t + 22 2 2 1 = − ln(cos22x + 4) + C. 2 t2 −5 3x + 5 ∫ dx . §Æt t = 3x + 5 , th× 2tdt = 3dx vµ x = 5) . 3 x Do ®ã: 3x + 5 t 2 dt dx = 2 ∫ 2 ∫ t −5 x t− 5 ⎛ 5⎞ 1 = 2 ∫ ⎜1 + 2 ⎟dt = 2t + 10. +C . ln t −5⎠ t+ 5 ⎝ 25
3x + 5 − 5 = 2 3x + 5 + 5 ln + C. 3x + 5 + 5 6) ∫ tg x.dx = ∫ [ tgx(1 + tg2x) - tgx].dx = ∫ tgxd (tgx ) − ∫ tgxdx 3 12 = tg x + ln cos x + C. 2 1 dx = (1 + tg2x)dx vµ do ®ã Chó ý cã thÓ ®Æt t = tgx, th× dt = 2 cos x dt dx = 1+ t 2 7) ∫ x a − x .dx . §Æt t = 3 2 a − x 2 , th× xdx = - tdt vµ x2 = a - t2. VËy: ∫x a − x 2 .dx 3 at 3 t3 1 =∫ (a − t 2 )t ( −tdt ) = ∫ (t 4 − at 2 ) dt = t 5 − + C = (3t 2 − 5a ) + C = 5 3 15 3 x + 2a 2 =− . (a − x 2 ) 3 + C . 15 x −x 2 8) ∫ dx . (x − 2) 3 §Æt t = x - 2, th× dx = dt vµ x2 = t2 + 4t + 4; x2 - x = t2 + 3t + 2. VËy: x2 − x t 2 + 3t + 2 ⎛1 3 2 ⎞ 3 1 ∫ (x − 2) 3 dx = ∫ dt = ∫ ⎜ + 2 + 3 ⎟dt = ln t − + 2 + C 3 t ⎝t t t⎠ t t 1 3 +C = ln x − 2 − + x − 2 (x − 2) 2 3x − 5 = ln x − 2 − +C. (x − 2) 2 x − 1 dx ∫ .. 9) x +1 x2 x −1 u2 +1 4udu x −1 2 §Æt u = , th× u = vµ do ®ã x = , dx = . (1 − u 2 ) 2 x +1 1− u2 x +1
4u du x − 1 dx (1 − u 2 ) 2 ∫ = ∫ u. = . VËy x +1 x2 2 ⎛ u2 +1⎞ ⎜1− u2 ⎟ ⎜ ⎟ ⎠ ⎝ 4u 2 du 4 4 ∫ (u 2 + 1) 2 = ∫ u 2 + 1 du − ∫ (u 2 + 1) 2 du ⎛1 u ⎞ 1 + arctgu ⎟ + C = 4arctgu - 4. ⎜ . 2 ⎝ 2 u +1 2 ⎠ 2u = 2arctgu − 2 +C u +1 x −1 x 2 −1 − +C = 2arctg x +1 x dx ∫ 10) . x − x2 2 ⎛1 ⎞ 1 ⎛1 ⎞ V×: x - x = − ⎜ − x ⎟ , nªn ®Æt t = ⎜ − x ⎟ th× dt = - dx vµ ta cã: 2 ⎝2 ⎠ 4 ⎝2 ⎠ dx dt ∫ x − x2 = −∫ = arccos2t + C = arccos(1 - 2x) + C. 2 ⎛1⎞ ⎜ ⎟ −t 2 ⎝2⎠ Còng cã thÓ ®Æt t = x , th× dx = 2tdt vµ x − x 2 = t 1 − t 2 . Do ®ã: dx 2tdt ∫ x − x 2 = ∫ t 1 − t 2 = 2arcsint + C = 2arcsin x + C. x +1 ∫ x(1 + xe dx . §Æt t = 1 + xex, th× dt = (x + 1)exdx, do ®ã: 11) x ) x +1 (x + 1)e x ∫ x(1 + xe x ) dx = ∫ xe x (1 + xe x ) dx = t −1 xe x dt =∫ = ln + C = ln +C t(t − 1) 1 + xe x t
dx ∫ . §Æt t = 1 + e x , th× exdx = 2tdt vµ ex = t2 - 1. 12) 1+ e x Do ®ã: t −1 dt 2tdt e x dx dx = 2∫ 2 ∫ t(t 2 − 1) = ln +C= ∫ ∫e = = t −1 t +1 1+ ex 1+ ex x 1 + ex −1 +C = ln 1+ ex +1 x2 −1 13) ∫ 4 dx . x +1 1 Chia c¶ tö sè vµ mÉu sè cho x2 vµ ®Æt t = x + , ta tÝnh ®−îc: x x2 −1 x 2 − 2x + 1 1 ∫ x4 +1 +C dx = ln 2 x + 2x + 1 22 x2 −1 14) I = ∫ 2 dx . (x + 5x + 1)(x 2 − 3x + 1) 1 C¸ch 1. Chia c¶ tö sè vµ mÉu sè cho x2 vµ ®Æt t = x + , ta tÝnh ®−îc: x ⎛ 1⎞ ⎜1 − 2 ⎟dx (x 2 − 1)dx 1⎛ 1 1⎞ dt ⎝ x⎠ = = =⎜ − ⎟dt ⎞ (t + 5)(t − 3) 8 ⎝ t − 3 t + 5 ⎠ (x + 1 + 5x)(x + 1 − 3x) ⎛ ⎞⎛ 2 2 1 1 ⎜ x + + 5 ⎟⎜ x + − 3 ⎟ ⎝ x x ⎠⎝ ⎠ VËy: 1 x 2 − 3x + 1 1 t −3 x2 −1 I=∫ 2 + C. + C = ln 2 dx = ln 8 t+5 8 x + 5x + 1 (x + 5x + 1)(x 2 − 3x + 1) C¸ch 2. V×: x2 + 5x +1 vµ x2- 3x +1, lµ c¸c tam thøc bËc hai nªn b»ng ph−¬ng ph¸p hÖ sè bÊt ®Þnh ®Æt: (x 2 − 1) Ax + B Cx + D =2 +2 , ta tÝnh ®−îc: (x + 1 + 5x)(x + 1 − 3x) x + 5x + 1 x − 3x + 1 2 2 2 5 2 3 A=− ;B=− ;C= ;D=− . 8 8 8 8
(x 2 − 1) 1 2x + 5 1 2x − 3 =− . 2 +.2 VËy: 2 . (x + 1 + 5x)(x 2 + 1 − 3x) 8 x + 5x + 1 8 x − 3x + 1 Tõ ®ã suy ra kÕt qu¶ nh− trªn. 3. Dïng ph−¬ng ph¸p tÝch ph©n tõng phÇn tÝnh c¸c tÝch ph©n sau: 1) ∫ x arctgxdx 2) ∫ (x + 2x + 3)cosxdx 2 2 3) ∫ x lnxdx 4) ∫ (x + 1)e dx − 2x 3 2 5) ∫ e cos3xdx 6) ∫ sin 3 x .dx − 2x arctgx arcsinx ∫x ∫ dx dx 7) 8) (1 + x 2 ) 2 x2 xe arctgx xln(x + 1 + x 2 ) ∫ ∫ dx dx . 9) 10) (1 + x 2 ) 3 1+ x2 Gi¶i. x3 dx 1) ∫ x arctgxdx . §Æt u = arctgx vµ dv = x dx th× du = 2 2 vµ v = , 1+ x 2 3 do ®ã: x3 1 x 3 dx ∫ x arctgxdx = 3 arctgx – 3 ∫ x 2 + 1 2 x3 1 x(x 2 + 1) − x arctgx – ∫ dx = x 2 +1 3 3 x3 1 ⎛ x2 1 ⎞ arctgx – ⎜ − ln(x + 1) ⎟ + C = 2 ⎜2 2 ⎟ = 3 3⎝ ⎠ x3 12 1 arctgx – x + ln(x + 1) + C . 2 = 3 6 6 2) ∫ (x + 2x + 3)cosxdx . 2 §Æt u = x2 + 2x + 3, dv = cosxdx th× du = (2x + 2)dx, v = sinx, do ®ã ∫ (x + 2x + 3)cosxdx = (x2 + 2x + 3)sinx – 2 ∫ (x + 1)sinxdx 2 = (x2 + 2x + 3)sinx + 2 ∫ (x + 1)d(cosx) = (x2 + 2x + 3)sinx + 2[(x + 1)cosx – ∫ cosxdx ]
= (x2 + 2x + 3)sinx + 2[(x + 1)cosx – sinx] + C = (x + 1)2sinx + 2(x + 1)cosx + C. 3) ∫ x lnxdx . §Æt u = lnx, dv = x3dx , tõ ®ã ta cã: 3 14 1 ∫ x lnxdx = x lnx − x4 + C. 3 4 16 4) ∫ (x + 1)e dx . §Æt u = x2 + 1 vµ dv = e– 2xdx th× du = 2xdx vµ − 2x 2 1 v = − e– 2x. 2 VËy: 1 ∫ (x 2 + 1)e − 2x dx = − e– 2x.(x2 + 1) + ∫ xe − 2x dx 2 ⎛ 1 − 2x ⎞ 1 = − e– 2x.(x2 + 1) + ∫ xd ⎜ − e ⎟ ⎝2 ⎠ 2 1 1 1 = − e– 2x.(x2 + 1) − xe– 2x + ∫ e dx − 2x 2 2 2 1 1 1 = − e– 2x.(x2 + 1) − xe– 2x − e– 2x + C 2 2 4 1 1 = − e– 2x.(x2 + x + ) + C. 2 2 5) ∫ e cos3xdx . − 2x 1 §Æt u = e– 2x vµ dv = cos3xdx th× du = - 2 e– 2xdx vµ v = sin3x. 3 1 – 2x 2 VËy: I = ∫ e cos3xdx = ∫e − 2x − 2x sin3xdx e sin3x + 3 3 1 – 2x 2 = e sin3x + J . 3 3 §Ó tÝnh J ta ®Æt u = e vµ dv = sin3xdx th× du = - 2 e– 2xdx vµ – 2x 1 v = − cos3x. 3 VËy 1 2 − 2x 1 2 J = − e– 2xcos3x − ∫ e cos3xdx = − e– 2xcos3x − I. 3 3 3 3
1 – 2x 2 1 2 e sin3x + J vµ J = − e– 2xcos3x − I Tõ hÖ I = 3 3 3 3 1 - 2x ta suy ra I = e (3sin3x - 2cos3x) + C. 13 Chó ý cã thÓ xÐt bµi to¸n tæng qu¸t: ∫ e cosbxdx . − ax 6) ∫ sin 3 x .dx . §Æt t = 3 x th× dx = 3t2dt vµ ta cã: I = ∫ sin 3 x .dx = 3∫ t sint.dt . 2 §Ó tÝnh tÝch ph©n nµy ta ®Æt u = t2 vµ dv = sintdt th× du = 2tdt vµ v = - cost. Do ®ã: ∫ t sint.dt = - t2cost + 2 ∫ tcost.dt = - t2cost + 2 ∫ td(sint) 2 = - t2cost + 2[tsint − ∫ sintdt ] = - t2cost + 2tsint + 2cost = (2 - t2)cost + 2tsint. Do ®ã: ∫ sin x 2 )cos 3 x + 6 3 x .sin 3 x + C. 3 3 x .dx = 3(2 - ⎛ 1⎞ 1 1 arcsinx 1 arcsinx.d ⎜ − ⎟ = − arcsinx + ∫ . ∫ ∫ x 2 dx = .dx 7) x 1− x2 x⎠ ⎝ x ⎛1⎞ d⎜ ⎟ ⎝ x ⎠ = − 1 arcsinx − ln 1 + 1 − 1 + C = 1 = − arcsinx + + ∫ x2 x x 2 x ⎛1⎞ ⎜ ⎟ −1 ⎝x⎠ 1+ 1− x2 1 = − arcsinx − ln +C. x x arctgx ∫ x 2 (1 + x 2 ) dx . 8) 1 dx dx , th× du = §Æt u = arctgx vµ dv = vµ x (1 + x ) 1+ x2 2 2 ⎛1 1⎞ 1 v = ∫⎜ 2 − dx = − − arctgx . 2⎟ 1+ x ⎠ x ⎝x VËy:
⎛1 ⎞ dx arctgx ⎛1 ⎞ dx = arctgx ⎜ − − arctgx ⎟ + ∫ ⎜ + arctgx ⎟ ∫ x 2 (1 + x 2 ) ⎠1+ x 2 ⎝x ⎝x ⎠ ⎛1 x⎞ ⎛1 ⎞ = arctgx ⎜ − − arctgx ⎟ + ∫ ⎜ − 2 ⎟dx + ∫ arctgx.d(arctgx) = ⎝ x x +1⎠ ⎝x ⎠ 1 1 1 = − arctgx - (arctgx)2 +ln x – ln(x2 + 1) + (arctgx)2 + C = x 2 2 x 1 1 − arctgx − (arctgx)2 + C. = ln 1+ x2 x 2 xe arctgx ∫ dx . 9) (1 + x 2 ) 3 §Ó tÝnh tÝch ph©n nµy ta ®Æt u = arctgx tøc lµ x = tgu. Khi ®ã: 3 3 du 1 (1 + x ) = (1 + tg u) = 2 2 2 2 3 , dx = . cos 2 u cos u Suy ra: xe arctgx tgu.e u .cos 3 u sinu − cosu u ∫ dx = ∫ .du = ∫ e u .sinu.du = .e + C . 2 cos u 2 (1 + x 2 ) 3 x 1 Nh−ng ta cã: sinu = ; cosu = , nªn suy ra: 1+ x2 1+ x2 xe arctgx x −1 ∫ dx = earctgx + C. (1 + x ) 23 2 1+ x 2 xln(x + 1 + x 2 ) ∫ dx . 10) 1+ x 2 xdx 1 + x 2 ) vµ dv = §Æt u = ln(x + . 1+ x2 dx 1+ x2 . Suy ra du = vµ v = 1+ x 2 Tõ ®ã: xln(x + 1 + x 2 ) ∫ dx = 1 + x 2 . ln(x + 1 + x 2 ) - x + C. 1+ x 2
4. TÝnh c¸c tÝch ph©n sau: x2 +1 x2 + x 2) ∫ 4 1) ∫ 6 dx dx x + x2 +1 x +1 x5 − x x 2 −1 4) ∫ 8 3) ∫ 4 dx dx x +1 x + x3 + x2 + x +1 x4 +1 1 ∫ x 6 x 2 + 1 dx 5) ∫ 6 dx ( ) 6) x +1 x 3n −1 1 8) ∫ 2n 7) ∫ 4 dx , (n lµ sè tù nhiªn). dx (x + 1) 2 x + x2 +1 cos 4 x dx ∫ 9) ∫ dx 10) sin 3 x sin 3 x.cos 5 x Gi¶i. 1) Ta cã: x2 + x dx 3 dx 2 1 1 1 1 dt ∫ x6 +1 dx = ∫ +∫ = arctgx 3 + ∫ 3 () () . 2 t +1 3 x3 2 +1 2 x2 3 +1 3 dt Ta tÝnh riªng tÝch ph©n: ∫ 3 . t +1 Ph©n tÝch: 1 t−2 1 ⎛ 1 2t − 1 ⎞ 1 3 11 11 1 =. − .2 =. − .⎜ . 2 − .2 ⎟. t +1 3 t + 1 3 t − t +1 3 t + 1 3 ⎝ 2 t − t + 1 2 t − t +1⎠ 3 V× vËy dt ∫ t3 +1 = ⎛ 1⎞ d⎜ t − ⎟ 1 1 1 1 1 ⎝ 2⎠ ln t + 1 − ln t 2 − t + 1 + ∫ = ln t + 1 − ln t 2 − t + 1 2 ⎛ 1 ⎞2 ⎛ 3 ⎞2 3 6 3 6 t− ⎟ +⎜ ⎟ ⎜ ⎝ 2⎠ ⎜ 2 ⎟ ⎝ ⎠ 2t − 1 1 + C. arctg + 3 3
VËy: ( )2 2x 2 − 1 x2 +1 x2 + x 1 1 1 ∫ x6 +1 + +C. dx = arctgx 3 + ln 4 arctg 12 x − x 2 + 1 2 3 3 3 x2 +1 2) ∫ 4 dx . x + x2 +1 ⎛ 1⎞ 1 Chia c¶ tö vµ mÉu cho x2, sau ®ã ®Æt t = x − ⇒ dt = ⎜1 + 2 ⎟dx , ®ång thêi ta x ⎝ x⎠ 1 còng cã x2 + 1 + 2 2 = t + 3. x VËy: ⎛ 1⎞ ⎜1 + 2 ⎟dx x +1 x2 −1 2 dt 1 t 1 ⎝ x⎠ = ∫ x 4 + x 2 + 1 dx = ∫ 2 ∫ t 2 + 3 = 3 arctg 3 + C = 3 arctg x 3 + C 1 x +1+ 2 x x 2 −1 3) ∫ 4 dx . x + x3 + x2 + x +1 ⎛ 1⎞ 1 Chia c¶ tö vµ mÉu cho x2, sau ®ã ®Æt t = x + ⇒ dt = ⎜1 − 2 ⎟dx , ®ång thêi x ⎝ x⎠ ta còng cã 2 11⎛ 1⎞ ⎛ 1⎞ x2 + x +1+ + 2 = ⎜ x + ⎟ − 2 + ⎜ x + ⎟ +1 = t 2 + t −1. xx x⎠ x⎠ ⎝ ⎝ VËy: ⎛ 1⎞ 1 5 d⎜ t + ⎟ t+ − x −1 2 dt 1 ⎝ 2⎠ 2 2 +C ∫ x 4 + x 3 + x 2 + x + 1 dx = ∫ t 2 + t − 1 = ∫ = ln 2 1 5 5 ⎛ 5⎞ 2 ⎛ 1⎞ ⎜ t+ + 2. ⎟ ⎜t + ⎟ −⎜ ⎝ 2⎠ ⎝ 2 ⎟ 2 2 2 ⎠ hay x 2 −1 2x 2 + x − 5x + 2 1 ∫ x4 + x3 + x2 + x +1 +C. dx = ln 5 2x 2 + x + 5x + 2 x5 − x 4) ∫ 8 dx . x +1
x5 − x x4 −1 1 t 2 −1 §Æt t = x , th× dt = 2xdx. Do ®ã ∫ 8 dx = ∫ 8 xdx = ∫ 4 2 dt . x +1 x +1 2 t +1 B»ng c¸ch chia c¶ tö vµ mÉu cho t2 vµ lµm t−¬ng tù nh− trªn ta ®−îc: x5 − x x 4 − 2x 2 + 1 1 ∫ x8 + 1 dx = +C . ln 4 x + 2x 2 + 1 42 x4 +1 5) ∫ 6 dx . ViÕt x4 + 1 = (x4 - x2 + 1) + x2, ta t¸ch thµnh hai tÝch ph©n: x +1 x4 +1 x4 − x2 +1 x 2 dx 1 ∫ x 6 + 1 dx = ∫ ( x 2 + 1)( x 4 − x 2 + 1) dx + ∫ ( x 3 ) 2 + 1 = arctgx + 3 arctgx + C . 3 1 ∫ x (x ) dx . 6) +1 6 2 1 dt §Æt x = ⇒ dx = − . t2 t VËy: t 6 dt t 6 dt 1 ∫ x 6 x 2 + 1 dx = − ∫ = −∫ ( ) . t2 +1 ⎛ 1⎞ t 2 ⎜1 + 2 ⎟ ⎝ t⎠ Ta ph©n tÝch hµm sè d−íi dÊu tÝch ph©n nh− sau: t6 +1−1 4 2 t6 1 =2 = t − t +1− 2 . t +1 t +1 t +1 2 Tõ ®ã ta cã: t5 t3 1 1 1 1 1 ∫ x 6 x 2 + 1 dx = − 5 + 3 − t + arctgt + C = − 5x 5 + 3x 3 − x + arctg x + C . ( ) 1 ∫x dx . Ta cã: 7) + x2 +1 4 1 ⎛ x +1 x −1 ⎞ 2x + 1 2x − 1 1 1 1 =⎜2 −2 ⎟= + − + x + x + 1 2 ⎝ x + x + 1 x − x + 1 ⎠ 4(x + x + 1) 4(x + x + 1) 4(x − x + 1) 4(x − x + 1) 4 2 2 2 2 2 1 x2 + x +1 1 1 x3 VËy: ∫ 4 dx = ln 2 + +C. arctg x + x2 +1 4 x − x +1 2 3 1− x2 a+b (Chó ý lµ ta cã c«ng thøc arctga + arctgb = arctg ). 1 − ab
x 3n −1 8) ∫ 2n dx ( n lµ sè tù nhiªn). (x + 1) 2 1 2n 1 n −1 3 n −1 §Æt xn = t, th× nx dx = dt ⇒ x dx = x dt = t 2 dt . n n VËy: x 3n −1 1 tdt ∫ ( x 2n + 1) 2 dx = n ∫ t. (t 2 + 1) 2 . §Ó tÝnh tÝch ph©n nµy ta dïng ph−¬ng ph¸p tÝch ph©n tõng phÇn b»ng c¸ch ®Æt nh− sau: ⎧ ⎧ du = dt u=t ⎪ tdt ⎪ 1 = dv ⇒ ⎨v = − 2 ⎨ . ⎪ (t + 1) ⎪ 2(t + 1) 2 2 ⎩ ⎩ Tõ ®ã: 1⎛ 1 dt ⎞ x 3n −1 1⎛ t ⎞ 1 tdt t ∫ ( x 2n + 1) 2 dx = n ∫ t. (t 2 + 1) 2 = n ⎜ − 2(t 2 + 1) + 2 ∫ t 2 + 1 ⎟ = − 2n ⎜ t 2 + 1 − arctgt ⎟ + C ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 ⎛ xn n⎞ = − ⎜ 2n ⎜ x + 1 − arctgx ⎟ + C . ⎟ 2n ⎝ ⎠ cos 4 x 9) ∫ dx = ∫ cos 3 x. sin −3 x.d (sin x ) . 3 sin x §Æt u = cos3x, dv = sin- 3xdx th× 1 −2 du = - 3cos2x.sinx.dx vµ v = − sin x . 2 Do ®ã: cos 4 x cos 3 x 3 cos 2 x ∫ sin 3 x dx = − 2 sin 2 x − 2 ∫ sin 2 x . sin x.dx cos 3 x 3 1 − sin 2 x 2 sin 2 x 2 ∫ sin x =− − dx cos 3 x 3 dx 3 −∫ + ∫ sin xdx = =− 2 2 sin x 2 sin x 2 cos 3 x 3 3 x − − ln tg − cos x + C 2 2 sin x 2 2 2
dx ∫ 10) sin 3 x.cos 5 x dx dx (sin 2 x + cos 2 x)dx dx + = = = Ta cã: sin 3 xcos 5 x sin 3 xcos5 x 3 5 3 5 sin xcos x sin xcos x sin 4 x cos 4 x 1 3 dx dx − = (tgx) 2 d(tgx) + (tgx) 2 d(tgx) . = + cos 2 x cotgx cos 2 x tg 3 x Tõ ®ã suy ra: dx 3 1 2 2 − ∫ sin 3 x.cos 5 x = 3 (tgx) − 2(tgx) 2 + C = 3 tgx tgx − 2 cotgx + C . 2 5. TÝnh c¸c tÝch ph©n sau: π π 1 + sinx 4 2 1) ∫ ln 2) ∫ ln (1 + tgx)dx dx 1 + cosx 0 0 1 1 dx 4) ∫ (arccotgx − arctgx)dx 3) ∫ x (e + 1)(x 2 + 1) −1 0 Gi¶i. π π π 1 + sinx 2 2 2 1) ∫ ln dx = ∫ ln(1 + sinx)dx − ∫ ln(1 + cosx)dx = I − J . 1 + cosx 0 0 0 π − t , thÕ th× dx = - dt, 1 + sinx = 1 + cost vµ ta cã: §Ó tÝnh I ta ®Æt x = 2 π π π 0 2 2 2 I = ∫ ln(1 + sinx)dx = ∫ ln(1 + cost)(−dt) = ∫ ln(1 + cost)dt = ∫ ln(1 + cosx)dx = J . π 0 0 0 2 Suy ra: π 1 + sinx 2 ∫ ln 1 + cosx dx = 0 0
π 4 π π π 2) I = ∫ ln (1 + tgx)dx . §Æt x = − t , th× khi x = 0 th× t = cßn khi x = th× 4 4 4 0 1 − tgt 2 ⎛π ⎞ t = 0, ®ång thêi 1 + tgx = 1 + tg ⎜ − t ⎟ = 1 + = vµ dx = - dt. 1 + tgt 1 + tgt ⎝4 ⎠ VËy: π π ⎛2⎞ 0 4 4 π I = ∫ ln (1 + tgx)dx = − ∫ ln⎜ ⎜ 1 + tgt ⎟dt = ∫ [ln 2 − ln(1 + tgt)]dt = ln2 − I . ⎟ 4 ⎠ ⎝ π 0 0 4 Tõ ®ã suy ra: π 4 π I = ∫ ln (1 + tgx)dx = ln2 . 8 0 1 dx ∫ (e 3) . + 1)(x 2 + 1) x −1 §Æt x = - t, th× khi x = - 1 ta cã t = 1, cßn khi x = 1 th× t = - 1, ®ång thêi et +1 x dx = - dt vµ e + 1 = . et Khi ®ã: −1 1 1 (e t + 1 − 1)dt e t dt dx ∫ (e x + 1)(x 2 + 1) = − ∫ (e t + 1)(t 2 + 1) = −∫1 (e t + 1)(t 2 + 1) −1 1 1 1 dt dt =∫ 2 −∫ t = t + 1 −1 (e + 1)(t 2 + 1) −1 ⎛ π⎞ π π 1 − I ⇒ 2I = ⎜− ⎟ = . = arctgt - −1 ⎝ 4⎠ 4 2 1 dx π ∫1 (e x + 1)(x 2 + 1) = 4 . VËy I = − 1 4) ∫ (arccotgx − arctgx)dx . 0 1 1 x π 1 21 Ta cã: I = ∫ arccotgx.dx = x.arcctgx 0 + ∫ 1 dx = + ln(1 + x ) 0 1+ x 2 4 2 0 0
π 1 = + ln2. 4 2 1 π 1 T−¬ng tù J = ∫ arctgx.dx = – ln2 . 4 2 0 1 VËy: ∫ (arccotgx − arctgx)dx = ln 2 0 6. TÝnh giíi h¹n sau: 1 ⎡ 12 n2 ⎤ 22 + + ... + lim ⎢ ⎥ n → ∞ n2 n +1 n+2 n + n⎦ ⎣ Gi¶i. ⎡⎛ 1 ⎞2 ⎛ 2 ⎞2 ⎛n⎞ ⎤ 2 ⎢⎜ ⎟ ⎜⎟ ⎜ ⎟⎥ 1 ⎡ 12 n 2 ⎤ 1 ⎢⎝ n ⎠ 22 ⎝ n ⎠ + ... + ⎝ n ⎠ ⎥ §Æt un = n 2 ⎢ n + 1 + n + 2 + ... + n + n ⎥ = n .⎢ n + 1 + n + 2 n+n⎥ = ⎣ ⎦ ⎢n n⎥ n ⎣ ⎦ ⎡⎛ 1 ⎞2 ⎛ 2 ⎞2 ⎛n⎞ ⎤ 2 ⎢⎜ ⎟ ⎜⎟ ⎜ ⎟⎥ ⎢ ⎝ n ⎠ + ⎝ n ⎠ + ... + ⎝ n ⎠ ⎥. 1 = ⎢ n⎥n 1 2 1+ 1+ 1+ ⎥ ⎢ n⎦ n n ⎣ x2 C¸c sè h¹ng trong tæng trªn lµ gi¸ trÞ cña hµm sè f(x) = t¹i c¸c ®iÓm 1+ x 1 2 n , ..., x n = = 1 . x1 = , x2 = n n n Râ rµng c¸c ®iÓm nµy chia ®o¹n [0, 1] thµnh n ®o¹n b»ng nhau víi 1 ®é dµi mçi ®o¹n lµ Δ x = . n 2 x Hµm sè f(x) = lµ kh¶ tÝch trªn ®o¹n [0, 1] nªn suy ra: 1+ x 1 1 x2 1 ⎡ 12 n2 ⎤ ⎛ 1⎞ 22 ⎥ = nlim u n = ∫ 1 + x dx = ∫ ⎜ x − 1 + x + 1 ⎟dx + + ... + lim 2 ⎢ ⎣n +1 n + 2 n + n⎦ 0⎝ ⎠ n→∞n →∞ 0 1 ⎛ x2 ⎞ 1 = ⎜ − x + ln( x + 1) ⎟ = ln 2 − ⎜2 ⎟ . 2 ⎝ ⎠ 0
7. Chøng minh r»ng: π 2 2 3 3 cotgx 1 2) ∫ (lnx) dx ≤ ∫ lnxdx ≤∫ dx ≤ 2 1) 12 π x 3 1 1 4 Gi¶i. ⎡π π ⎤ cotgx trªn ®o¹n ⎢ , ⎥ ta cã 1) XÐt hµm sè f(x) = x ⎣4 3⎦ 1 − .x − cot gx ⎛π π ⎞ x + sin x cos x sin 2 x ==− < 0 , víi mäi x ∈ ⎜ , ⎟ . f'(x) = 2 2 2 ⎝4 3⎠ x sin x x ⎡π π ⎤ VËy f(x) nghÞch biÕn trªn ⎢ , ⎥ , do ®ã: ⎣4 3⎦ 3 cotgx 4 ⎛π ⎞ ⎛π π ⎞ ⎛π⎞ ≤ ≤. f ⎜ ⎟ ≤ f(x) ≤ f ⎜ ⎟ , víi mäi x ∈ ⎜ , ⎟ , tøc lµ: π π x ⎝3⎠ ⎝4⎠ ⎝4 3⎠ Tõ ®ã suy ra: π π 4 ⎛π π ⎞ 3 ⎛π π ⎞ 3 3 cotgx 3 cotgx 1 ∫x ≤∫ dx ≤ ⎜ − ⎟ hay ⎜ − ⎟≤ dx ≤ . π ⎝3 4⎠ π ⎝3 4⎠ 12 π x 3 π 4 4 2 2 2) ∫ (lnx) dx ≤ ∫ lnxdx . 2 1 1 Víi mäi x ∈ (1, 2) th×: ln1 < lnx < ln2 < lne ⇔ 0 < lnx < 1 ⇒ (lnx)2 < lnx. Tõ ®ã: 2 2 ∫ (lnx) dx ≤ ∫ lnxdx . 2 1 1 8. Chøng minh r»ng nÕu hµm sè f''(x) liªn tôc trªn ®o¹n [a, b] th×: b ∫ xf ′′ (x)dx =[bf ′ (b) − f(b)] − [af ′ (a) − f(a)] a Gi¶i. C¸ch 1. XÐt hµm sè F(x) = xf'(x) - f(x), ta cã: F'(x) = f'(x) + xf''(x) - f'(x) = xf''(x). Nh− vËy F(x) lµ mét nguyªn hµm cña xf''(x). Do ®ã:
b ∫ xf ′′ (x)dx = (xf ′( x) − f ( x)) = [bf ′ (b) − f(b)] − [af ′ (a) − f(a)] . b a a C¸ch 2. Dïng tÝch ph©n tõng phÇn. b b b ∫ xf ′′ (x)dx = ∫ xd[f ′ (x)] = xf ′ (x) − ∫ f ′( x)dx = bf ′ (b) − f(b) − f ( x) b = b a a a a a = [bf ′ (b) − f(b)] − [af ′ (a) − f(a)] 9. Chøng minh r»ng nÕu hµm sè f(x) liªn tôc trªn ®o¹n [a, b] th×: b 1 ∫ f(x)dx =(b − a)∫ f [a + (b − a)x]dx a 0 Gi¶i. §Æt x = a + (b - a)t, th× dx = (b - a)dt, khi x = a th× t = 0 vµ khi x = b th× t = 1. VËy: b 1 1 ∫ f(x)dx = ∫ f [a + (b − a)t](b - a)dt = (b − a)∫ f [a + (b − a)t]dt = a 0 0 1 = (b − a)∫ f [a + (b − a)x]dx 0 10. Chøng minh r»ng: π π π 2 π ∫ xf(sinx)dx = 2∫ f(sinx)dx = π ∫ f(sinx)dx 0 0 0 Gi¶i. §Æt x = π - t th× π 0 π π ∫ xf(sinx)dx =∫ (π − t)f(sint)(−dt) = π ∫ f(sint)dt − ∫ tf(sint)dt 0 π 0 0 π π π Suy ra ∫ xf(sinx)dx = 2∫ f(sinx)dx . 0 0 §Ó tÝnh tÝch ph©n thø hai nµy ta t¸ch thµnh hai tÝch ph©n nh− sau: π π π 2 ∫ f(sinx)dx = ∫ f(sinx)dx + ∫ f(sinx)dx . π 0 0 2 π π §Æt x = π - t th× khi x = th× t = , cßn khi x = π th× t = 0. 2 2