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Tính nguyên hàm bằng phương pháp từng phần ðịnh lý : Nếu u, v là hai hàm số liên tục trên K thì... Nguyễn Phú Khánh Ờ đà Lạt.[r]

Nguyễn Phú Khánh Ờ đà Lạt lại ∫ ( ) P x (x − a )(x − b )(x − c ) A = P( x) + c + D ⇒ A, B,C dx P (x ) ðặt : (x − a )(x − b ) ∫ (x − α )(x − β ) ∫ x − a dx = = dx = A B C + + x −a x −b x −b ( ∫ (x − a )(x + a ) dx = x − 2dx = ∫ (x − 2)2 d (x − 2) = ⇒ A, B,C , D  1   x −a  ln − dx =   2a ∫  x − a x + a  2a  x + a    +C   (x − 2)2 + C ( x ∫ xe dx = ) (x − b )  1  x −α − +C ln  dx = ∫ α − β x −α x − β  α −β x −β dx −2 ∫ (x + 3)2 = ∫ (x + 3) d(x + 3) = − x + ∫ B + (x − a )(x − b )(x − c ) (x − a ) (x − b ) (x − c ) ∫ (x − a )(x − b ) ∫ dx P (x ) ðặt : http://www.maths.vn ) −1 +C x2 x2 e d x = e +C ( ) 2∫ − +1 1 (1 + x ) = ∫ (1 + x ) d(1 + x ) = 3 1 + x3 − +1 x dx − 3 + C = (1 + x ) + C 2  −  2 3 x + dx = x dx + x dx = x + x + C = x + 33 x + C 2  ∫  ∫ ∫  3 x    1 ∫  cos x - sin x - 2x + x  dx = 3∫ cos x dx + 2∫ - sin x dx − ∫ 2x dx + ∫ x dx = sin x + cos x − x + ln x + C ( ) ( ) ( ) cos 8x cos 2x + ) +C   sin x dx ∫ cos2 x dx = ∫  cos2 x + cos x −  dx = tan x − 2x + ∫ + ∫ cos 2xd 2x = tan x − x + sin 2x + C cos xdx ∫ cotx − sin 2x +  dx = ∫ sin x − ∫ sin 2x + d 2x + d sin x 3 =∫ − ∫ sin 2x + d 2x + = ln | sin x | + cos(2x + 1) + C sin x 2 ∫ sin 3x cos 5xdx = ∫ (sin 8x − sin 2x )dx = (− ( ) ( ) ( ( ) ( ) Lop12.net ) ( ) Nguyễn Phú Khánh Ờ đà Lạt ∫ (3 cos x − ∫ x −1 dx = + ex http://www.maths.vn x 3 dx = sin x − + C = sin x − +C ∫ 3 ln ln d + ex  = x − ln + e x + C dx = ∫ dx − ∫ x +e  )dx = ∫ cos xdx −  ex −  + ex  ∫ x −1 x ( ) Tính nguyên hàm các hàm số sau : f (x ) = x − ( ) f x = ( x + 1)(x − x + 1) x ( ) f x =( f (x ) = 2x + 3x 1 − )(x − x ) x x ( ) ( ) f x = cos 2x + sin 3x + x ( ) f x = sin x + cos x + x + x Tìm nguyên hàm F ( x ) hàm số π  f x = cos 2x + sin 3x + x biết F   = −3 3 x3 − x2 − biết F −3 = 10 f x = x2 ( ) ( ) ( ) Tính nguyên hàm phương pháp ñổi biến số ðịnh lý : Cho hàm số u = u ( x ) có ñạo hàm liên tục trên K và hàm số y = f (u ) liên tục cho f u (x )  xác ñịnh trên K Khi ñó F là nguyên hàm f , tức là ∫ f (u )du = F (u ) + C thì ∫ f u (x )dx = F u (x ) + C I = ∫x ðặt : t = I = ∫x dx x2 + x + ⇒ t = x + ⇒ td t = x d x dx xdx td t = ∫ = ∫ = t −1 t x2 + x2 x2 + ( I = 1 t −1   x2 + −   +C ln ln  + C =    t +   x2 + + 1   I = ∫ xdx + x2 + (1 + x ) ) ∫ Lop12.net dt = 2 t −1  1  −  ∫  t − t +  d t Nguyễn Phú Khánh Ờ đà Lạt ðặt : t = http://www.maths.vn xdx + x ⇒ t = x + ⇒ tdt = xdx ⇒ = + x2 + + x2 I = ∫ dt 1+t tdt t 1+t = = + t + C = + + x2 + C Tính nguyên hàm các hàm số sau ∫ (5x + 3) dx ∫ (2x + 1) dx 20 2 ∫ (1 + x ) xdx 9x ∫ 1-x ∫ x (1 + x )2 ∫ dx − 4x + ∫ x − 6x + 9dx dx ∫ 2x + ∫x ∫x ( ∫x ) dx ∫ x ln5 x − x dx 34 − x dx dx (8x + 27) 3 3x + dx + 6x + 2x + ∫ x + x − 2x dx 3x + x + ∫ x + dx x + 4x + ∫ x + dx x3 ∫ x + 2x + dx 3x + 3x + ∫ x − 3x + dx 3x + ∫ x − 2x − 5x + dx dx − 5x + ln x + ∫ x dx dx ∫ x ln x ln(ln x ) x 2dx x + 4x ∫1 x dx ∫x ∫x dx ∫x − x2 dx π ∫ e x + 1dx e 2x ∫ e 2x + dx ex ∫ − e 2x dx dx ∫ ∫ sin x sin 4x dx ∫e ∫e cosx sin xdx s inx cos xdx ∫e + e 2x Lop12.net cos x sin xdx dt 1+t Nguyễn Phú Khánh Ờ đà Lạt ( ) dx ln ex http://www.maths.vn ∫e ∫ + x ln x x cos x dx − 4e −x ∫ sin x dx ∫ cos xdx ∫ ( tan x + tan x )dx 4 ∫ e x + e −x + 2dx + sin x ∫ − cos xdx ∫ sin ∫ cos ∫ sin ∫ sin ∫ sin x cos2 xdx x cos5 xdx x cos2 xdx x x ∫ sin 3cos dx ∫ sin x cos xdx sin x ∫ cos4 x dx sin x ∫ cos6 x dx ∫ t anxdx ∫ cot xdx ∫ (tan x + t anx )dx ∫ (cot x + cot x )dx dx ∫ sin x ± cos x cos 2x ∫ + sin x cos xdx tan x ∫ cos x dx ∫ cot x ∫ + sin2 x dx sin x + cos x dx x sin x  ∫  tan x + dx ∫ cos ∫ cos  dx 2x + + 2x −  x dx x x dx Tính nguyên hàm phương pháp phần ðịnh lý : Nếu u, v là hai hàm số liên tục trên K thì ∫ xdx sin x ∫ cos5 x dx I = ∫ sin ∫ xdx ∫ + cos x dx sinx ∫ cos4 x dx I = ∫ u(x )v '(x )dx x ln(x + x + 1) dx x2 + x ln(x + x + 1) dx = x2 + ∫ ln(x + x + 1) x x2 + Lop12.net dx = u(x )v(x ) − ∫ v(x )u '(x )du Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn  x u = ln x + x + 1+   x + dx = du = ðặt :  ⇒  x x + x2 + dx dv =  x2 + v = x +   ) ( ( I = x + ln x + ) ∫ xdx x2 + − = dx x2 + ( x + ln x + ) x2 + − x +C Tính nguyên hàm các hàm số sau ∫ x s inxdx ∫ (2x + 1) sin xdx ∫ xe -x ∫ (x + 1)e dx ∫e s inxdx x x dx ∫ x ln xdx ∫e ∫e x ∫x x ∫ x sin 3dx cos x + ∫ x + dx sin 2xdx ln xdx x +1 dx ∫ x tan xdx ∫ x tan xdx ∫ ln(x + x )dx ∫ x ln 1−x dx 1+x Tích Phân Ví dụ : Lop12.net ∫ x 2dx (1 − x ) Nguyễn Phú Khánh Ờ đà Lạt ∫2 ) ∫d( dx = x x = x = 42−43 ∫ ∫ x − 3dx = 2  = 16 ( x − 3) x −  3  x − 3d (x − 3) = ∫ 25 − 3x d (25 − 3x ) dx = http://www.maths.vn ∫ −3 = 25 − 3x −2 −2 25 − 3x = ( 13 − 5) 3 ( ) d x2 + x + 1 2x + ln dx = = x + x + = ln ∫0 x + x + ∫0 x + x + 1 5xdx ∫ (1 − x ) 1+x =− d 1−x ∫0 − x ( ∫ 1−x dx = ∫ (−1 + ( 2 dx ∫3 x − 3x + = ) ) = − (1 − x ) −2 2 − x2 = −2 ( 5 ) = 35 36 ( (x − 1) − (x − ) dx =  −1 +  dx = ln x − ∫ (x − 1)(x − ) ∫  x −1 x −  x −1   3 π π π 4 sin x dx = tan xdx = ∫ cos x −π ∫π − −d (cos x ) = − ln cosx cosx π   ∫π cot xdx = π∫  sin2 x −  dx = −cotx − x ( ) ln + 20 ln − 10 ln ) π π =1− = π −π =0 π π  3 − cot 2x  dx = ∫π cos2 x ∫π  cos2 x − sin2 x  dx = tan x + 2cotx ( π π  dx  ∫π sin2 x cos2 x = π∫  sin2 x + cos2 x  dx = tan x − cotx ( ) ) π π π π = 6 −x ∫ x e dx = 4 π = ln ( − π = − − ln 2 dx  1  x −3 = ∫ − = ln  dx = ln x − −1 2 − 4x + −1  x − x −  −1 1 2xdx dx dx = + = ln x + + ln x − = ln ∫0 x − ∫0 x + ∫0 x − 1 20dx  3x +  dx ∫0 x − 5x − dx =  ∫0 x + + ∫0 x −  = ln x + + 20 ln x − ∫π 4 ∫x ) 1 )dx = − ∫ dx + ∫ dx = − ∫ dx − ∫ d − x = −x − ln − x 1−x 1−x 1−x 2 2 −1 − x −x 2 e d ( x ) e − = = (1 − e −9 ) ∫ −2 2 Lop12.net = 11 −5 3 ( ) Nguyễn Phú Khánh Ờ đà Lạt 2 e xdx ∫ x= −1 + e −1 e2 e2 ∫ d (e + 2) x + ex = ln (2 + e x ) = ln −1 dx d (ln x ) ∫e x ln x = ∫e ln x = ln ln x e http://www.maths.vn e2 e e −1 −2 −2 −1 −5 − 2x + x ∫ x + dx = ∫ ( −x − 1) dx + ∫ ( ∫ 1−x −1 ∫ ∫ ∫ dx = = 1 Ngoài , ta có thể ñặt t = ln x −1 ) −5 dx = 1−x −1 e  x2   x2  x + dx =  − − x  +  + x  =   −2   −1 (1 − x )2 −5 2e + = ln ln x (ln x )5 = (ln ) (ln ) = dx x d x ∫1 x ∫1 2e + e 1−x −5 ∫ 1−x 1−x ∫ − x dx = x dx = −1 −1 −1 −5 =4  x3  x3 x − 1dx = ∫ (1 − x )dx + ∫ (x − 1)dx =  x −  + ( − x ) =   1 2 2 2 π π π ∫ cos x dx = ∫ cos xdx + π∫ ( − cos x ) dx = sin x 0 π − sin x π π =2 2 2π ∫ 2π − cos 2xdx = ∫ 2π π  sin x dx =  ∫ sin xdx − ∫ sin xdx  =   π 0  π π ∫  x3 x2   x3 x2  49 x − x − dx = ∫ (−x + x + 6)dx + ∫ (x − x − 6)dx =  − + + 6x  +  − − 6x  = 2 3  0  3 + cos 2x dx = π 0 π ∫ cos x dx = ∫ cos xdx − π∫ cos xdx = 2 I = ∫π − sin 2x (2 + sin x ) dx ðặt : t = sin x ⇒ dt = cos xdx x = ⇒ t =  ðổi cận  π x = − ⇒ t = −1  Lop12.net Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn  + t) − ( I = 2∫ = 2∫ dt = ∫  − 2 + t (2 + t ) (2 + t ) (2 + t )  0 tdt  d + t 2  2 −1 −1 −1 ( )   I =  ln + t +  = ln − 2 + t  −1  ( ∫x ) dx x2 + x ðặt : t = x + ⇒ dt = x2 + dx = xdx tdt ⇒ dx = t x  x = ⇒ t = ðổi cận  x = ⇒ t = dx x x2 + = tdt x2 x2 + = (t tdt ) −1 t = 1 1  dt = −   dt t2 −  t − t +   1  ∫ x x + ∫2  t − − t +  dt = ln t − − ln t + ∫x ( dx ) 3 1 t −1 =  ln  = ln 2 t + 12 2 dx +x +1  1 x + x + = x +  + 2  ðặt x + =    π π  1 3 dt = tan t  t ∈  − ;   ⇒ dx = d  tan t −  = + tan2 t dt   2 cos t  2    (  π ⇒t = x = ⇒ tan t = ðổi cận  x = ⇒ tan t = ⇒ t = π  Lop12.net ) Nguyễn Phú Khánh Ờ đà Lạt ( ( Vậy dx ∫0 x + x + = ∫ π ∫x ) ) π + tan2 t dt 3 dt = = π ∫ 3 π + tan t π http://www.maths.vn dx + 4x + 1 1 1  = = −    x2 + x2 +  x + 4x + x2 + x2 + ( )( ) 1 dx  dx dx  = −  ∫0 x + 4x +  ∫0 x + ∫0 x +  = A + B   1 A= dx ∫0 x + ( ) + tan2 t dt −π dx π < t < ⇒ dx = + tan t dt ⇒ = = dt ðặt : x = tan t, 2 x +1 + tan2 t ( ) x = ⇒ t =  ðổi cận  π x = ⇒ t =  π A = ∫ dt = t π = π B= dx ∫0 x + ðặt : x = ( ) x = ⇒ t =  ðổi cận  π x = ⇒ t =  π B= ∫x ( ) + tan t dt −π π dx 1 dt tan t, < t < ⇒ dx = + tan2 t dt ⇒ = = 2 x +1 tan t + 3 ∫ dt = π t = π dx 1π π  = −   + 4x +   Lop12.net Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn I = ∫ x x − a dx , a > 0 −x ax + •a ≥ ñó I = − ∫ x (x − a )dx = = a − a  −x ax   x ax  + − •0 < a < ñó I = − ∫ x (x − a )dx + ∫ x (x − a )dx =   +    a   a a a3 a3 a a3 a3 a3 a + + − − + = − + 3 3 I =− Cho ( ) ∫ f x dx = −2 và ( ) ( ) ∫ g x dx = Hãy tính ( ) ( ) ( ) ( ) ( ) ∫ 3f x − g x  dx, ∫ 5 − f x  dx 1 ( ) ( ) ( ) I = ∫ 3 f x − g x  dx = ∫ f x dx − ∫ g x dx = ∫ f x dx − ∫ g x dx = −9   1 3 ( ) ( ) J = ∫ 5 − f x  dx = ∫ 5dx − ∫ f x dx = 5x   1 1 + = 23 5 Cho ∫ f (x )dx = −4, ∫ f (x )dx = 6, ∫ g(x )dx = Hãy tính : ∫ f (x )dx , ∫ 4 f (x ) − g(x ) dx 1 5 ∫ f (x )dx + ∫ f (x )dx = ∫ f (x )dx ⇔ 5 2 1 ∫ f (x )dx = ∫ f (x )dx − ∫ f (x )dx ⇔ 5 1 ∫ f (x )dx = 10 ∫ 4 f (x ) − g(x ) dx = ∫ f (x )dx − ∫ g(x )dx = 16 1 ∫ (x + x + 1)dx ∫ (x + x x )dx ∫ 0 x +1 ∫2 x dx x +1 ∫1 x dx ∫( 2 x3 ∫0 x + 1dx ∫x −1 e ∫−2 (11 + 5x )3 dx (1 + x + 1dx x + 1)(x − x + 1)dx 1 )dx x ∫( x − 1)(x + x + 1)dx 1 ∫1 (x + x + x + x )dx ∫ (x + x x + x )dx ∫ Lop12.net dx x +1 + x −1 Nguyễn Phú Khánh Ờ đà Lạt dx ∫−1 x − 3x + 2 ∫ (x ) + 2x − dx x3 + 1 http://www.maths.vn x dx ∫2 (1 − x )9 (2x − 10x + 16x − 1).dx ∫ x − 5x + −1 (x − 3x + x + 6).dx ∫ x − 5x + 6x −1 x 3dx ∫2 (x − 1)10 dx ∫ x ( x + 1) −1 ∫x −1 (7x − 4)dx − 3x + ∫x dx − 4x + dx ∫ (x + 3) (x + 1) 2 ∫ 1 ∫ dx x + x2 5 dx ∫x x x2 + 4 ∫ x 2 dx ∫ x x +9 2 dx x2 − dx x2 − ∫x 2dx 4x − dx 3 ∫ x − dx ∫0 x − 3x + 2dx ∫ x − x dx ∫ ∫( 1 ∫0 x x − dx x − 6x + dx x +1 x ∫x −1 dx − x − 12 ∫ ∫π − sin x dx π ∫ cos x sin xdx x − 2x + xdx ∫ ) x + − x − dx −3 π 4x − x − 2x − 1dx π ∫ cos x e ∫ ln x dx e Lop12.net sin xdx Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn π ∫x + x − 2dx ∫ π −3 tan2 x + cot2 x − 2dx π ∫ sin x − cos x dx ∫x ( ) − a + x + a dx − 2x + m dx ∫x 1 ∫ (e + x )dx x ∫e e x ln ∫ (e + x + 1)dx x ∫ 0 ex ex − dx ex + x −1 dx π π ∫ (2 sin x − cos x ).dx π ∫ sin 2x (1 + sin x ) −π dx π ∫ sin 2x cos 3x dx π 2 ∫π (3 sin x + 2cosx + x )dx ∫ sin x sin 2x sin 3xdx π π ∫ ( sin 2x ) + cos 2x dx π π ∫ ( sin ) ∫ π sin x x + cos x dx cos x dx x ∫ π sin Tính tích phân phương pháp ñổi biến số b Công thức dổi biến số ∫ f [u(x )]u ' (x ) dx = a u (b ) ∫ f (u )du u (a ) Lop12.net dx + cos x Nguyễn Phú Khánh Ờ đà Lạt e ∫x I = dx − ln2 x ðặt t = ln x ⇒ dt = dt ∫ I = − t2 = dx x = ⇒ t = 0, x = e ⇒ t = x π 1 x dx ∫0 − x ( http://www.maths.vn ∫ ) ∫ x x + dx ∫ x − x dx ∫ x + x dx x dx 1 + x2 x2 ∫ x3 + dx 1 ∫ x x + 1dx x + 1.dx 0 ∫x x + 1dx x + 1dx ∫x − x dx ∫x 15 + 3x dx ∫ x − xdx 1 ∫ x ∫ 2x + x dx ∫ dx ∫ 2x + ∫ ( x + 1) x + 2x + 4.dx 0 x + dx x −1 x +1 −3 ∫x −8 dx ∫x dx (1 + x ) dx 1−x 2a − x dx (a > 0) Lop12.net x2 + ∫3 ∫ x3 + x + 2x 3 x3 + dx dx x5 + ∫ x4 ∫ x3 + 2a ∫ x −1 xdx ∫ −1 − x dx ∫x ∫x 2 ∫ ∫x dx x −3 dx x +1 +x +3 (x − 1).dx x +1 Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn a ∫x a − x dx (a > 0) ∫ xdx ( x + 1) ∫1+ 10 x x −1 ∫ dx ∫x −2 ∫ x −1 −1 ∫ 1 ∫ ∫ ∫ x − 4x x +1 + x ∫ dx ∫ dx 1+x −1 −1 x +5 +4 −1 dx ∫ x (1 + x +1 − x dx dx x + + x +1 dx ∫ 2x + + x −4 + x +2 xdx ∫ 2+x + 2−x ∫3 x +1 −1 π π 4 x −3 dx x +1 +x +3 π ∫ tan xdx ∫ tan π π π 2 ∫ x dx + sin xcosxdx ∫ cotxdx π ∫π sin2 x cos2 x dx ∫ π sin x 6 π π sin x ∫0 + cos xdx 2 ∫ cos 2x ( sin ) x − cos4 x dx sin2008 x ∫π cos2012 xdx π dx ∫ cos + cos x π dx π ∫ sin 2x (1 − sin x )dx π 4x + 2 dx x) x +1 dx x ∫ dx ∫1+x + 2dx ∫ dx dx x +1 x x + 1dx x +2 1 ∫x dx x dx ∫1+ 2x + 1 dx x ∫ x Lop12.net cos x ( + sin x ) dx Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn π π π 2 ∫ sin xcos xdx sin x π ∫ sin xcos xdx π ∫ + 3cosx dx 3 π π π − sin x ∫π sin2 x dx − sin2 x ∫0 + sin 2x dx π π π 4 0 π π π π π π 2 ∫ sin x cos2 2x dx ∫ cos x sin x dx sin 2x ∫ + sin π π tan x ∫0 + cos xdx cot x ∫π sin2 x + 1dx π cos 2x ∫ + sin 2xdx sin x cos x ∫0 + cos2 x dx sin 2x ∫ + cos x dx sin 4x ∫0 + cos2 xdx xdx cos x ∫ + sin xdx ∫ cos 0 ∫ sin sin 3x ∫ + cos xdx 4 x .dx dx x + cos4 x π π π sin 2x cos x ∫0 + cos x dx − sin2 x ∫0 + sin 2x dx π π π 2 ∫ ∫ sin 2x + sin x + cos x cos x + cos x dx dx ∫ cos 2 ∫ x cos 4xdx 0 π π π cos 2x dx ∫0 + cos x sin x cos3 x ∫0 cos2 x + dx ∫ π π π 2 2 sin 2x dx ∫0 cos4 x + ∫ cos x cos 2xdx π π sin 2x dx x ∫ + sin sin x ∫0 sin x + cos x dx π ∫ π ∫ sin x cos x (1 + cos x ) dx sin x − cosx + sin 2x dx dx ∫π sin x + cos x .dx sin 2x cos2 x + sin2 x ∫ π − + cos x x cos3 xdx sin 2x + sin x ∫ sin π − cos3 x sin x cos5 xdx π π ∫ π π sin xdx ∫0 sin x + cos x sin 2x cos2 x + sin2 x + π ∫ π cos x cos xdx ∫0 sin x − cos x Lop12.net dx tan x + cos2 x dx dx Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn π  π sin  x −  dx 4  ∫0 sin 2x + 2(1 + sin x + cos x ) π + sin 2x + cos 2x ∫π sin x + cos x dx π π 2 ∫π e sin x cosxdx ∫π e cosx ∫e sin xdx x +2 xdx ln ∫ e x − 1dx dx ∫ 2 ∫ 1+x dx 1−x ∫ − x dx −1 2 ∫ x2 + x + dx ∫ − 4x dx (x + 1)(x + 2) ∫x ∫x − x dx 2 0 ∫ 2 ∫ 1 ∫ x −1 dx x 1 − − x2 ∫0 + x dx x dx x2 + dx 4x + 8x )dx 1 ∫−1 x + 2x + 2dx (4x + 11).dx + 5x + ∫x 1 ∫0 + x 2dx + 2x + + x + ∫x + 2 (x + 2) ∫2 (x + 1)2 dx 1 − x2 dx x2 ∫ (x ( x + − 2)dx 1 ∫ (1 + 3x 2 ) x ∫ (1 + 2x ) dx 3x ∫0 x + 2x + dx .dx Lop12.net x + 3x + 10 ∫0 x + 2x + dx Nguyễn Phú Khánh Ờ đà Lạt ∫ http://www.maths.vn dx + x2 ∫ ∫ x dx ∫0 x + ∫ (x + 2x + 10x + 1).dx ∫0 x + 2x + dx x2 + x dx ∫0 x + x + 2x − dx + x2 ∫x 3 − x + 2x + 10x + ∫0 x + 2x + dx 3x + dx + x2 1 dx +x +1 x ∫ (x − 1) (x + 1) dx 1+ dx x − 2x + 10 ∫ 1+ ∫ x2 + dx x4 − x2 + x7 ∫2 + x − 2x 4dx ∫x (1 − x ).dx x (x + 1) ∫x 2 ∫ 4 dx ∫1 x (x + 1)2 ∫x 1 dx + 4x + dx + 4x + dx + 2x + x x dx ∫0 (x − 4)2 3.dx ∫0 x + (x − x − 4x − 1).dx ∫1 x4 + x3 2x + 2x + 13 ∫0 (x − 2)(x + 1)2 dx x 5dx ∫ x6 − x3 − 3 e + ln x ∫1 x dx e ∫ e + ln x dx x ∫x e ∫ − ln2 x +1 x ) dx e2 ∫ e + ln x ln x dx x ∫x 1 ∫ cos (1 + ln x ) dx e e dx (2 ln x − ln x + ln x dx Lop12.net Nguyễn Phú Khánh Ờ đà Lạt e ∫ ln x + ln x dx x e ∫x ln x ln x ∫x dx ln x + 1 (x + 1).dx + x ln x ∫x dx ( ln x + 1) http://www.maths.vn e3 e e ln x +1 ∫1 x dx e sin(ln x ) ∫1 x dx e2 ln 1 + ln2 x ∫e x ln x dx dx ∫ln e x + 2e −x − ∫0 + 2x dx ln 1 ∫ e x + 1.e 2xdx ln dx ∫0 e 2x + e x dx ∫0 + e x ∫e e −x ∫0 e −x + dx ln ln ln ∫e x +e x ∫ ∫ e x − 1.dx e (1 + e x )2 ∫0 + e 2x dx e 2x dx ex + ln ln ex ∫ (e x dx ex + ∫ dx dx +3 2x +1 ) dx π π π cos x ∫0 − sin x + sin2 x dx dx ∫0 + sin 2x ∫ + cos x dx π π π 2 cos x ∫0 − sin x − cos2 xdx ∫0 + sin x + cos x dx ∫ cos 2x ( sin π π π dx ∫π sin x cos x ∫ π π 4 sin x ( sin x + cos x ) ∫e π π 0 ( sin x + cos x + ) ) x + sin x dx ∫ cos 2x ( sin x − cos x + ) Lop12.net dx cos x π cos 2x sin2 x ∫ sin x + dx dx cos2 x ∫ sin x + 5π π π ∫ ( cos sin x cos xdx cos 2x ∫0 sin x + cos x + 2dx ∫ sin2 x π sin 2x ∫0 + cos4 xdx tan x dx ∫π cos2 x − sin x cos x ) x + cos x dx dx cos 2x ∫π cos x − dx cos x sin x dx Nguyễn Phú Khánh Ờ đà Lạt http://www.maths.vn π π π + sin 2x + cos 2x ∫π sin x + cos x dx sin x + cos x + ∫0 sin x + cos x + dx 3 ∫ π 6 π ∫ cos 2x (cos dx  π sin x sin  x +  6  x + sin x ).dx π ∫ ( cos ) x + sin10 x − cos4 x sin x dx 10 Tính tích phân phương pháp phần b b b Công thức tích phân phần : ∫ u(x )v '(x )dx = u(x )v(x ) a − ∫ v(x )u '(x )dx a Dạng u = f (x ) du = f '(x )dx   sin ax  sin ax      ⇒    dv = cos ax  dx v = ∫ cosax  dx eax  e ax           dx u = ln(ax ) du = ðặt  ⇒ x dv = f (x )dx v = ∫ f (x )dx   sin ax  β   f x cosax ( )  dx ðặt ∫ α e ax    β Dạng 2: a ∫ f (x )ln(ax )dx α Ví dụ : e A = ∫ x ln xdx I = ∫ x ln xdx 1 π B = ∫ x e dx x J = ∫ x sin xdx 0 e A = ∫ x ln xdx  dx du = u = ln x  x ðặt:  ⇒ dv x dx x =  v =  e e e x3 e3 x ln x − ∫ x 2dx = − A = ∫ x ln xdx = 31 1 e = e e − 2e + − = 9 B = ∫ x 2e xdx Lop12.net Nguyễn Phú Khánh Ờ đà Lạt u = x ðặt  http://www.maths.vn du = 2xdx ⇒  x x v = e dv = e dx B = ∫ x 2e xdx = x 2e x u = x ðặt  1 0 − ∫ xe xdx = e − 2C ,C = ∫ xe xdx du = dx ⇒ dv = e dx v = ex    x1 x  B = e −  xe − ∫ e dx  = −e + e x   0   x =e −2 I = ∫ x ln xdx  du = dx x ⇒ x v =  u = ln x ðặt:  dv = x 4dx  5  x5  x5 1 x5 I =  ln x  − ∫ dx = 625 ln − ∫ x 4dx = 625 ln − 51 5  1 x = 625 ln − 3124 25 π J = ∫ x sin xdx u = x ðặt:  du = dx ⇒ v = − cos x  dv = sin xdx  π J = −x cos x π π 2 ( ) − ∫ − cos x dx = ∫ cos xdx = sin x 0 π =1 ln I = ∫ xe −x ( ) J = ∫ x ln + x dx dx 0 ln I = ∫ xe −x dx u = x ðặt:  −x dv = e dx  I = −xe −x ln ( du = dx ⇒ v = −e −x  ln − ∫ ( −e ) dx = − ln 2.e −x ln − ln − ( ) −x ∫ e d −x ( ) = − ln − e −x ) J = ∫ x ln + x dx Lop12.net ln = 1 − ln 2 2