bài tập tích phân có giải

Chuyên đề: Nguyên hàm – Tích phân 2 1 0 2 xdx I x = + ∫ 3 2 1 1 7 x x dx I x − = − ∫ 6 3 2 1 2 1 4 1 I dx x x = + + + ∫ ( ) 4 4 3 2 1 dx I x x = + ∫ 10 5 5 2 1 dx I x x = − − ∫ 1 8 3 6 0 1I x x= − ∫ 1 7 0 3 2 2 1 1 x I dx x + = + + ∫ 1 2 8 2 0 4 2 5 2 x x I dx x x + = + + ∫ ( ) 2 2 9 1 1 2 2 x I dx x x − = + + ∫ 2 2 5 2 10 0 1I x x dx= + ∫ 11 2 0 sin 3 cos x x I dx x π = + ∫ ( ) 2 1 12 2 0 1 1 x I dx x + = + ∫ ( ) 3 2 13 3 2 1 1 x x I dx x − + = − ∫ 2 14 0 sin sin cos x I dx x x π = + ∫ 3 3 15 4 6 sin cos x I dx x π π = ∫ 3 16 3 4 1 cos .sin I dx x x π π = ∫ ( ) 2 3 3 17 0 sin cosI x x dx π = + ∫ ( ) 2 18 2 0 sin 2 2 sin x I dx x π = + ∫ ( ) 2 19 2 0 sin 2 2 sin x I dx x π = + ∫ ( ) 4 20 2 0 cos2 sin cos 2 x I dx x x π = + + ∫ ( ) 2 3 2 21 0 1 sin sinI x xdx π = − ∫ 3 2 22 0 4sin 1 cos x I dx x π = + ∫ 3 2 23 2 0 sin cos 1 cos x x I dx x π = + ∫ ln3 24 0 2 x dx I e = + ∫ 1 2 25 0 1 x x e I dx e = − ∫ 2 26 0 .sin .cosI x x xdx π = ∫ ( ) 27 2 1 ln 1 e e x x I dx x + = + ∫ 10 2 28 1 lgI x xdx= ∫ ( ) ( ) 2 29 0 2 5 ln 1I x x dx= + + ∫ ( ) 1 2 2 30 2 0 1 x x e I dx x = + ∫ ( ) 1 2 31 0 ln 1I x x dx= + ∫ ( ) 2 32 sin 2 x I x e x dx − = + ∫ 2 33 2 2 cos 4 sin x x I dx x π π − + = − ∫ 4 sin 34 0 (tg cos ) x I x e x dx π = + ∫ 2 35 1 3 ln ln 1 e x I dx x x = + ∫ 36 1 3 2ln 1 2ln e x I dx x x − = + ∫ 4 37 0 2 1 1 2 1 x I dx x + = + + ∫ 2 38 0 sin 2 3 4sin cos2 x I dx x x π = + − ∫ 1 3 2 39 2 0 4 x x I xe dx x   = −  ÷ −   ∫ 2 40 2 2 0 sin sin 2 3sin 4 x x I dx x cos x π + = + ∫ ( ) 0 3 41 1 1 x I x e x dx − − = + + ∫ ( ) ln3 43 3 0 1 x x e I dx e = + ∫ ( ) 2 1 3 2 44 0 1 x I x e x dx= + + ∫ 4 45 0 1 cos 2 x I dx x π = + ∫ 1 Lời Giải: 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 0 2 6 t dt I t ′ ⇒ = = − ∫ ( ) 2 2 2 0 0 6 6 2 1 2 3ln 2 2 3ln(2 3) 6 6 t dt t t t   −   = + = + = + −  ÷  ÷  ÷ − +     ∫ 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= − ⇒ = − ⇒ = − ( ) ( ) 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   = − + =  ÷   2 7, 1 7 0 3 2 2 1 1 x I dx x + = = + + ∫ (đổi biến 2 1 1t x= + + ) 8, ( ) ( ) 2 1 1 4 14 1 14 9 2 2 ln 2 ln 3 8 2 3 2 3 2 1 3 2 0 0 2 5 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 +   = = + = + + = +  ÷ + +   ∫ ∫ 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 −   = = − = + = −  ÷   ∫ ∫ 3 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 ln3 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 3 3 2 2 2 2 0 0 0 0 cos sin 4 sin cos cos sin cos sin 3 3 3 x x xd x xd 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 = = + + ∫ ∫ ( ) ( ) ( ) ( ) 2 1 1 2 2 0 0 0 2 2 2 2 2 2 ln 2 2ln 2 1 2 2 2 t t dt dt t t t t + −   = = = + + = −  ÷ +   + + ∫ ∫ 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 2 2 2 4 0 0 0 0 2 3 5 0 1 cos 2 sin 2 1 cos cos 1 2cos cos cos 2 2 4 2cos cos 8 cos 4 3 5 4 15 x x x dx x d x x x d x x x x π π π π π π π −   = + − = − + − +  ÷     = + − + = −  ÷   ∫ ∫ ∫ 4 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 ln3 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   = = − =  ÷ +   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 10 10 2 2 2 1 1 1 1 10 2 2 1 1 2 1 1 100 lg 50 lg 50 lg lg 2 ln10 2ln10 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= + + = + + ∫ ∫ 5 ( ) ( ) ( ) 2 2 2 2 2 0 0 0 2 2 0 5 4 5 ln 1 14ln 3 4 1 1 14ln 3 4 4ln 1 18ln3 10 2 x x x x x dx x dx x 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 1 1 2 2 2 2 2 1 2 2 2 2 2 2 0 0 0 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 e e d x e xe dx xd e xe e dx x x 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 1 cos 2 2 cos 2 cos 2 2 2 1 1 1 1 1 cos 2 2 sin 2 cos 2 2 sin 2 sin 2 2 2 2 2 2 1 1 1 2 1 cos2 sin 2 cos 2 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 − − − − − − − − − = − − = − + − + = − − − + = − − − + + − = − + + − + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 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 − =− − − = = = − = − ⇒ = − − − − ∫ ∫ ∫ 6 ( ) 2 2 2 2 2 2 2 cos 1 1 1 1 2 sin ln 3 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 t dt t dt t t   − ⇒ = = − = − = −  ÷   ∫ ∫ 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 cos 2 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   = − = + −  ÷ − −   ∫ ∫ ∫ 7 4 1 4 3 2 2 2 2 2 2 3 0 3 1 1 4 1 1 1 2 1 1 32 61 8 6 3 3 3 2 2 2 2 2 2 2 3 4 2 3 4 12 x x e t e e e xe dt t t t       − +   = − − = + − − = − − = + −  ÷  ÷  ÷  ÷         ∫ 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= ⇒ = + ( ) ( ) 2 6 6 6 6 2 2 0 0 0 0 3 1 tan sin 1 1 sin 1 ln ln3 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 − = = − = − − = − + − ∫ ∫ 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 − − − − − = + + = + + = + ∫ ∫ ∫ ( ) ( ) 3 1 0 0 0 0 1 7 4 1 0 3 3 3 1 1 1 1 1 0 0 9 2 1, 1 3 1 3 7 4 28 t x x x x x t t A xe dx xd e xe e dx e B x x dx t t dt = + − − − − − − − − −   = = − = − + = − = + = − = − = −  ÷   ∫ ∫ ∫ ∫ ∫ Vậy 41 37 2 28 I A B e= + = − 2 2 2 43 3 2 2 2 2 2 2 2 2 1 tdt dt I t t t ⇒ = = = − = − ∫ ∫ 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= + ⇒ = + ⇒ = 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 dx 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 cos 2 2cos 2 2 x x I dx dx xd x x x xdx x x π π π π π    ÷ = = = = −  ÷ +   ∫ ∫ ∫ ∫ 8 4 0 1 1 2 1 ln cos ln ln 2 8 2 8 2 2 8 4 x π π π π = + = + = − 9 . dx e = + ∫ ( ) 2 1 3 2 44 0 1 x I x e x dx= + + ∫ 4 45 0 1 cos 2 x I dx x π = + ∫ 1 Lời Giải: 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 . Chuyên đề: Nguyên hàm – Tích phân 2 1 0 2 xdx I x = + ∫ 3 2 1 1 7 x x dx I x − = − ∫ 6 3 2 1 2 1 4 1 I dx x x = + + + ∫ (. 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 − =− −