Đề luyện tập số 3: Chuyên đề nguyên hàm – tích phân 1, ( ) ( ) 2 2 2 3 2 1 0 0 0 2 2 8 2 2 4 2 2 1 3 3 2 2 xdx I x dx x x x x = = + − = + − + = − ÷ ÷ + + ∫ ∫ 2, 3 3 3 3 3 2 1 1 1 1 1 ( 7) ( 1) ( 1) 1 6 1 1 ( 1) 6 7 7 7 7 x x dx x dx x x dx x dx I x d x x x x x − − − − − = = + = − − − − − − − ∫ ∫ ∫ ∫ ∫ ( ) 3 3 2 2 2 1 2 4 2 1 6 6 3 3 x I I ′ ′ = − − = − với 3 2 1 ( 1) 7 x dx I x − ′ = − ∫ Để tính 3 2 1 ( 1) 7 x dx I x − ′ = − ∫ ta đặt 2 1 1x t x t− = ⇒ = + ( ) 2 2 2 2 2 2 2 0 0 0 2 6 6 2 1 2 3ln 2 2 3ln(2 3) 6 6 6 t dt t I dt t t t t − ′ ⇒ = = + = + = + − ÷ ÷ ÷ − − + ∫ ∫ Do đó: 2 32 2 48ln(2 3) 3 I = − − 3, 6 3 2 1 2 1 4 1 I dx x x = + + + ∫ Đổi biến 2 4 1 4 1 2t x t x tdt dx= + ⇒ = + ⇒ = 5 5 5 3 2 2 3 3 3 ( 1) ( 1) 2 1 ( 1) ( 1) tdt d t d t I t t t t + + ⇒ = = − + + + + ∫ ∫ ∫ ( ) 5 3 1 3 1 ln 1 ln 1 2 12 t t = + + = − ÷ ÷ + 4, ( ) ( ) ( ) ( ) 4 3 3 4 4 4 3 4 3 3 2 2 2 2 1 1 1 1 1 1 65 ln ln 1 ln 2 ln 3 1 3 3 9 1 x dx d x dx I x x x x x x + − + = = − = − + = − + + ∫ ∫ ∫ 5, 10 5 5 2 1 2ln 2 1 dx I x x = = − − + ∫ (đổi biến 1t x= − ) 6, 1 8 3 6 0 1I x x= − ∫ Đổi biến 3 2 3 2 1 1 2 3t x t x tdt x dx= − ⇒ = − ⇒ = − Hocmai.vn – Ngôi trường chung của học trò Việt ( ) ( ) 0 1 2 2 2 6 4 2 6 1 0 2 2 1 2 3 3 I t t dt t t t dt⇒ = − − = − + ∫ ∫ 1 6 5 3 0 2 2 16 3 7 5 3 315 t t t = − + = ÷ 7, 1 7 0 3 2 2 1 1 x I dx x + = = + + ∫ (đổi biến 2 1 1t x= + + ) 8, ( ) ( ) 1 1 2 8 2 0 0 4 14 1 14 9 2 2 ln 2 ln3 2 5 2 3 2 3 2 1 3 2 x x I dx dx x x x x + = = − + = + − ÷ ÷ + + + + ∫ ∫ 9, ( ) ( ) ( ) ( ) 2 2 2 2 9 3 1 1 2 2 4 2 3 1 2 2 2 x x x I dx dx x x x + − + + − = = + + + ∫ ∫ ( ) ( ) ( ) 2 1 1 3 2 2 2 1 2 2 3 2x x x dx − − = + − + + + ∫ ( ) ( ) ( ) 2 3 1 1 2 2 2 1 2 5 2 2 2 6 2 2 3 3 3 x x x − = + − + − + = − 10, ( ) 2 3 2 2 3 7 5 3 1 2 5 2 2 2 10 0 1 1 2 1 1 7 5 3 t x t t t I x x dx t t dt = + = + = − = − + = ÷ ∫ ∫ 11, 11 2 0 sin 3 cos x x I dx x π = + ∫ Đổi biến t x dt dx π = − ⇒ = − ( ) ( ) ( ) ( ) ( ) 11 11 2 2 2 0 0 0 sin sin cos 3 cos 3 cos 3 cos t t t t d t I dt dx I t t t π π π π π π π π − − − ⇒ = = = − − + − + + ∫ ∫ ∫ ( ) ( ) 2 1 2 6 3 tan 11 2 2 2 0 1 6 3 1 tan cos 2 3 cos 2 3 2 3 3tan 6 3 u v v du d t du I t u v π π π π π π π = − − + − ⇒ = = = = + + + ∫ ∫ ∫ 12, ( ) ( ) 2 1 1 1 2 12 2 2 0 0 0 1 2 1 ln 1 ln 2 1 1 1 x x I dx dx x x x x + = = + = + + = + ÷ + + ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt 13, ( ) ( ) ( ) ( ) ( ) 3 2 3 3 2 13 3 3 2 2 2 2 1 1 1 1 1 1 5 ln 1 ln 2 1 4 1 1 2 1 x x x x I dx dx x x x x x − + − + − + = = = − − − = + − − − − ∫ ∫ 14, 2 14 0 sin sin cos x I dx x x π = + ∫ ( ) 2 0 sin cos ' 1 1 2 sin cos 4 x x dx x x π π + = − = ÷ + ∫ 15, 3 2 3 3 3 15 4 4 3 6 6 6 sin 1 cos 1 1 14 26 3 cos cos cos 3cos cos 3 27 x x I dx d x x x x x π π π π π π − = = − = + = − ÷ ∫ ∫ 16, ( ) ( ) 3 3 3 16 3 2 3 2 3 4 4 4 sin 1 cos cos .sin cos .sin 1 sin sin d x x I dx dx x x x x x x π π π π π π = = = − ∫ ∫ ∫ ( ) 3 3 2 2 3 2 2 3 2 2 2 2 1 1 1 1 dt t dt t t t t t = = + + ÷ − − ∫ ∫ 3 2 2 2 2 2 1 1 1 1 ln ln 1 ln 3 2 2 2 3 t t t = − − − = + ÷ 17, ( ) 2 2 2 3 3 3 3 17 0 0 0 sin cos sin cosI x x dx xdx xdx π π π = + = + ∫ ∫ ∫ ( ) ( ) 2 2 2 2 0 0 2 2 3 3 0 0 sin cos cos sin cos sin 4 cos sin 3 3 3 xd x xd x x x x x π π π π = − + = − − + − = ÷ ÷ ∫ ∫ 18, ( ) ( ) ( ) 2 2 18 2 2 0 0 sin 2 sin 2 sin 2 sin 2 sin x x I dx d x x x π π = = + + ∫ ∫ ( ) ( ) ( ) ( ) 1 1 2 2 0 0 2 0 2 2 2 2 2 2 2 2 ln 2 2ln 2 1 2 t t dt dt t t t t + − = = + + = + + = − ÷ + ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt 20, ( ) ( ) ( ) ( ) 4 4 20 2 2 0 0 sin cos cos sin cos2 sin cos 2 sin cos 2 x x x x x I dx dx x x x x π π + − = = + + + + ∫ ∫ Đặt sin cos 2 cos sint x x dt x x = + + ⇒ = − , khi 0 3; 2 2 4 x t x t π = → = = → = + Do đó: 2 2 2 2 2 2 20 2 2 3 3 3 2 1 2 2 2 2 5 ln ln 2 3 3 t I dt dt t t t t t + + + − + = = − = + = + − ÷ ÷ ∫ ∫ 21, ( ) 2 2 2 3 2 2 5 21 0 0 0 1 sin sin sin sinI x xdx xdx xdx π π π = − = − ∫ ∫ ∫ ( ) ( ) 2 2 2 2 0 0 2 2 2 4 0 0 2 3 5 0 1 cos 2 1 cos cos 2 sin 2 1 2cos cos cos 2 4 2cos cos 8 cos 4 3 5 4 15 x dx x d x x x x x d x x x x π π π π π π π − = + − = − + − + ÷ = + − + = − ÷ ∫ ∫ ∫ 22, ( ) 3 2 2 2 22 0 0 4sin 4sin cos 1 cos 1 cos x x I dx d x x x π π = = + + ∫ ∫ ( ) ( ) 2 2 2 0 0 cos 4 1 cos cos 4 cos 2 2 x x d x x π π = − = − = − ÷ ∫ 23, ( ) 3 2 2 2 2 23 2 2 0 0 sin cos 1 cos 1 cos 1 cos 2 1 cos x x x I dx d x x x π π = = − + + + ∫ ∫ ( ) 2 2 1 1 1 1 1 1 ln 2 ln 2 2 2 t dt t t t − − = = − = ∫ 24, ( ) ( ) ( ) ln3 ln3 ln 3 24 0 0 0 1 1 1 2 2 2 2 x x x x x x x d e dx I d e e e e e e = = = − ÷ + + + ∫ ∫ ∫ ln3 0 1 1 3 1 1 6 ln ln ln ln 2 2 2 5 2 2 5 x x e e = = − = ÷ + Hocmai.vn – Ngôi trường chung của học trò Việt 25, ( ) ( ) 1 1 1 2 25 0 0 0 1 1 1 1 1 1 x x x x x x x x e e I dx d e e d e e e e = = = − + − ÷ − − − ∫ ∫ ∫ ( ) ( ) ( ) 1 3 1 2 2 0 2 2 1 2 1 1 2 3 3 x x e e e e = − + − = − + ÷ 26, ( ) 2 3 3 3 26 0 0 0 0 1 1 .sin .cos cos cos cos 3 3 I x x xdx xd x x x xdx π π π π = = − = − − ÷ ∫ ∫ ∫ ( ) ( ) 3 2 0 0 1 1 sin 1 sin sin sin 3 3 3 3 3 3 x x d x x π π π π π = + − = + − = ÷ ∫ 27, ( ) ( ) 27 2 1 1 ln 1 ln 1 1 e e e e x x I dx x x d x x + = = − + ÷ + + ∫ ∫ ( ) ( ) 1 1 1 1 ln ln 1 1 e e e e x x d x x x x = − + + + ÷ + + ∫ 1 1 1 1 1 2 1 2 1 1 1 1 1 1 e e e e e e dx dx e x x e x e − − = − + + + = + = ÷ + + + + ∫ ∫ 28, ( ) ( ) 10 10 10 10 2 2 2 2 2 2 2 28 1 1 1 1 1 1 lg lg lg lg 2 2 I x xdx xd x x x x d x = = = − ÷ ∫ ∫ ∫ ( ) ( ) 10 10 2 1 1 10 10 2 2 1 1 10 2 2 1 1 2 1 100 lg 50 lg 2 ln10 2ln10 1 50 lg lg 2ln10 50 1 50 99 50 50 ln10 2ln 10 ln10 4ln 10 x xdx xd x x x x d x xdx = − = − ÷ = − − ÷ = − + = − − ∫ ∫ ∫ ∫ 29, ( ) ( ) ( ) ( ) 2 2 2 29 0 0 2 5 ln 1 ln 1 5I x x dx x d x x= + + = + + ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt ( ) ( ) ( ) 2 2 2 2 0 0 2 0 2 2 0 5 5 ln 1 1 4 14ln3 4 1 14ln3 4 4ln 1 18ln3 10 2 x x x x x dx x x dx x x x x + = + + − + = − + − ÷ + = − + − + = − ÷ ∫ ∫ 30, ( ) 1 1 2 2 2 2 30 2 0 0 1 1 1 x x x e I dx x e d x x = = − ÷ + + ∫ ∫ ( ) ( ) 1 1 1 2 2 2 2 2 2 0 0 0 1 1 2 2 1 2 2 2 0 0 0 1 2 2 2 2 0 1 2 1 1 2 2 2 1 1 2 2 2 2 2 2 x x x x x x x x e e d x e xe dx x x e e xd e xe e dx e e e e = − + = − + + + = − + = − + − = − = − − = ÷ ∫ ∫ ∫ ∫ 31, ( ) ( ) ( ) 1 1 2 3 31 0 0 1 ln 1 ln 1 3 I x x dx x d x= + = + ∫ ∫ ( ) ( ) 1 1 3 1 3 2 0 0 0 1 3 2 0 1 1 ln 2 1 1 ln 1 1 3 3 1 3 3 1 ln 2 1 2ln 2 5 ln 1 3 3 3 2 3 18 x x x dx x x dx x x x x x x = + − = − − + − ÷ + + = − − + − + = − ÷ ∫ ∫ 32, ( ) 2 2 2 32 sin 2 sin 2 x x I x e x dx x e dx x xdx − − = + = + ∫ ∫ ∫ ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 cos2 2 1 2 cos 2 cos 2 2 1 1 cos 2 2 sin 2 2 2 1 1 1 cos 2 2 sin 2 sin 2 2 2 2 1 1 1 2 1 cos 2 sin 2 cos2 2 2 4 x x x x x x x x x x d e x d x x e xe dx x x x xdx x e x x xd e xd x x e x x xe e dx x x xdx x x e x x x x x − − − − − − − − − = − − = − + − + = − − − + = − − − + + − = − + + − + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt 33, 2 2 2 33 2 2 2 2 2 2 cos cos 4 sin 4 sin 4 sin x x x x I dx dx dx x x x π π π π π π − − − + = = + − − − ∫ ∫ ∫ Ta có: ( ) 2 2 2 2 2 2 2 2 2 0 4 sin 4 sin 4 sin t x x t t A dx dt dt A A x t t π π π π π π − =− − − = = = − = − ⇒ = − − − − ∫ ∫ ∫ ( ) 2 2 2 2 2 2 2 cos 1 1 1 1 2 sin ln3 sin ln 4 sin 4 2 sin 2 sin 4 2 sin 2 x x B dx d x x x x x π π π π π π − − − + = = − + = − = − ÷ − − + − ∫ ∫ Vậy 33 ln3 2 I A B= + = − 34, 4 4 4 sin sin 34 0 0 0 sin (tan cos ) cos cos x x x I x e x dx dx e xdx x π π π = + = + ∫ ∫ ∫ ( ) 4 2 sin sin 4 4 2 0 0 0 2 ln 2 ln cos ln 1 2 2 x x x d e e e π π π = − + = − + = + − ∫ 35, 2 35 1 3 ln ln 1 e x I dx x x = + ∫ Đặt 2 1 ln 1 ln 1 2t x t x tdt dx x = + ⇒ = + ⇒ = ( ) ( ) ( ) ( ) 2 2 2 3 35 1 1 1 2 2 1 2 2 2 2 1 1 1 3 3 t I tdt t d t t t − ⇒ = = − − = − = ∫ ∫ 36, 36 1 3 2ln 1 2ln e x I dx x x − = + ∫ Đặt 2 1 2ln 1 2ln 1t x t x tdt dx x = + ⇒ = + ⇒ = ( ) 2 2 2 2 3 2 36 1 1 1 4 10 2 11 4 4 3 3 3 t t I tdt t dt t t − ⇒ = = − = − = − ÷ ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt 37, 4 37 0 2 1 1 2 1 x I dx x + = + + ∫ Đặt ( ) ( ) 2 1 2 1 1 2 1 1t x t x dx t dt= + + ⇒ − = + ⇒ = − ( ) 4 4 4 2 37 2 2 2 1 1 1 2 2 ln ln 2 2 2 t t I t dt t dt t t t t − ⇒ = − = − + = − + = − ÷ ÷ ∫ ∫ 38, ( ) 2 2 38 2 0 0 sin sin sin 2 3 4sin cos2 2 4sin 2sin xd x x I dx dx x x x x π π = = + − + + ∫ ∫ ( ) ( ) 1 1 2 0 0 1 1 1 1 2ln 2 1 ln 1 ln 2 2 1 2 2 4 2 1 tdt dx t t t − = = + + = − = ÷ ÷ + + ∫ 39, ( ) ( ) 1 1 1 3 2 2 2 2 39 2 2 0 0 0 1 1 4 2 2 4 4 x x x x I xe dx xd e d x x x = − = + − ÷ − − ∫ ∫ ∫ 4 1 4 3 2 2 2 2 3 0 3 2 2 1 1 4 1 1 1 2 8 2 2 2 2 2 2 2 3 1 1 32 61 6 3 3 3 4 2 3 4 12 x x e t e xe dt t t t e e − = − − = + − − ÷ ÷ ÷ + = − − = + − ÷ ∫ 40, 2 2 2 40 2 2 2 2 2 2 0 0 0 sin sin 2 sin sin 2 3sin 4 3sin 4 3sin 4 x x x x I dx dx dx x cos x x cos x x cos x π π π + = = + + + + ∫ ∫ ∫ Có: ( ) 1 2 2 2 2 2 2 0 0 0 cos sin 3sin 4 3 3 d x x dt A dx x cos x cos x t π π = = − = + + + ∫ ∫ ∫ Đặt ( ) 2 3 tan 3 1 tant u dt u du= ⇒ = + thì: ( ) ( ) 2 6 6 6 6 2 2 0 0 0 0 3 1 tan sin 1 1 sin 1 ln ln 3 cos 1 sin 2 1 sin 2 3 3tan u du d u du u A u u u u π π π π + + ⇒ = = = = = − − + ∫ ∫ ∫ ( ) ( ) 2 2 2 2 2 2 2 2 0 0 0 4 sin sin 2 2 4 sin 2 2 3 3sin 4 4 sin d x x B dx x x cos x x π π π − = = − = − − = − + − ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt Vậy ( ) 40 ln3 2 2 3 2 I A B= + = + − 41, ( ) 0 0 0 3 3 41 1 1 1 1 1 x x I x e x dx xe dx x x dx A B − − − − − = + + = + + = + ∫ ∫ ∫ Ta có: ( ) 0 0 0 0 1 1 1 1 2 1 x x x x A xe dx xd e xe e dx e − − − − − − − − = = − = − + = − ∫ ∫ ∫ ( ) 3 1 0 1 7 4 1 3 3 3 1 0 0 9 1 3 1 3 7 4 28 t x t t B x x dx t t dt = + − = + = − = − = − ÷ ∫ ∫ Vậy 41 37 2 28 I A B e= + = − 43, ( ) ln3 43 3 0 1 x x e I dx e = + ∫ Đặt 2 1 1 2 x x x t e t e tdt e dx= + ⇒ = + ⇒ = 2 2 2 43 3 2 2 2 2 2 2 2 2 1 tdt dt I t t t ⇒ = = = − = − ∫ ∫ 44, ( ) 2 2 1 1 1 3 2 3 3 2 44 0 0 0 1 1 x x I x e x dx x e d x x x dx A B= + + = + + = + ∫ ∫ ∫ Ta có: ( ) ( ) 2 2 2 2 1 1 1 1 3 2 2 2 0 0 0 0 1 1 2 2 x x x x A x e dx x d e x e e d x = = = − ÷ ∫ ∫ ∫ 2 1 0 1 1 1 2 2 2 2 2 2 x e e e e = − = − − = ÷ ÷ ( ) 2 2 1 2 5 3 1 3 2 2 2 0 1 1 2 2 2 1 1 5 3 15 t x t t B x x dx t t = + + = + = − = − = ÷ ∫ ∫ Vậy 44 1 2 2 2 17 4 2 2 15 30 I A B + + = + = + = 45, ( ) 4 4 4 4 4 45 2 0 0 0 0 0 1 1 tan tan tan 1 cos2 2cos 2 2 x x I dx dx xd x x x xdx x x π π π π π ÷ = = = = − ÷ + ∫ ∫ ∫ ∫ Hocmai.vn – Ngôi trường chung của học trò Việt 4 0 1 1 2 1 ln cos ln ln 2 8 2 8 2 2 8 4 x π π π π = + = + = − Hocmai.vn – Ngôi trường chung của học trò Việt . Đề luyện tập số 3: Chuyên đề nguyên hàm – tích phân 1, ( ) ( ) 2 2 2 3 2 1 0 0 0 2 2 8 2 2 4 2 2 1 3 3 2 2 xdx I x dx x x x