1. Chương II: Nguyên hàm và tích phân − Tr n Phương
BÀI 8. PHƯƠNG PHÁP TÍCH PHÂN T NG PH N
I. CÔNG TH C TÍCH PHÂN T NG PH N
Gi s u = u ( x ) ; v = v(x) có o hàm liên t c trong mi n D, khi ó ta có:
∫ ∫ ∫
• d ( uv ) = udv + vdu ⇔ d ( uv ) = udv + vdu ⇔ uv = udv + vdu ∫ ∫
b b
b
⇒ ∫ udv = uv − ∫ vdu ⇒ ∫ udv = ( uv )
a
a ∫
− vdu
a
Nh n d ng: Hàm s dư i d u tích phân thư ng là tích 2 lo i hàm s khác nhau
Ý nghĩa: ưa 1 tích phân ph c t p v tích phân ơn gi n hơn (trong nhi u
trư ng h p vi c s d ng tích phân t ng ph n s kh b t hàm s dư i d u tích
phân và cu i cùng ch còn l i 1 lo i hàm s dư i d u tích phân)
Chú ý: C n ph i ch n u, dv sao cho du ơn gi n và d tính ư c v ng th i
tích phân ∫ vd u ơn gi n hơn tích phân ∫ udv
II. CÁC D NG TÍCH PHÂN T NG PH N CƠ B N VÀ CÁCH CH N u, dv
1. D ng 1:
u = P ( x )
 sin ( ax + b ) dx 
 cos ( ax + b ) dx   sin ( ax + b ) dx
cos ( ax + b ) dx
∫ P (x) 
 e
ax + b ⇒
dx  dv =  ax + b
(trong ó P(x) là a th c)
 ax + b  e dx
 m dx   ax + b
  m dx
2. D ng 2:
dv = P ( x ) dx
 arcsin ( ax + b ) dx 
 arccos ( ax + b ) dx  arcsin ( ax + b )
 arccos ( ax + b )
arctg ( ax + b ) dx  
∫ P (x)  ⇒ arctg ( ax + b )
( ax + b ) dx u = 
(trong ó P(x) là a th c)
 arc cotg 
 arc cotg ( ax + b )
ln ( ax + b ) dx  ln ( ax + b )

 log m ( ax + b ) dx 



log m ( ax + b )
 
3. D ng 3:
 sin ( ln x )  eax+b sin ( αx + β) dx  eax+b
sin ( ln x ) dx  cos ( ln x )
 ( )  ax+b u =  ax+b
k cos ln x dx u = sin ( log x )  e cos ( αx + β) dx ⇒  m
∫ x 

⇒
sin ( loga x) dx  cos ( log x)

a ; ∫  max+b sin ( αx + β) dx 
dv = sin ( αx + β) dx
cos ( loga x ) dx   k  ax+b cos ( αx + β) dx
a

dv = x dx m cos ( αx + β) dx 
  
210

2. Bài 8. Phương pháp tích phân t ng ph n
III. CÁC BÀI T P M U MINH H A:
∫ P ( x ) {sin ( ax + b ) ; cos ( ax + b ) ; e ; m ax+b } dx
ax+b
1. D ng 1:
∫
• A1 = x 3 cos x dx .
u = x 3
 du = 3x 2 dx

Cách làm ch m: t  ⇒ . Khi ó ta có:
dv = cos x dx  v = sin x
 
u = x 2
 
du = 2x dx
∫
3 2
A1 = x sin x − 3 x sin x dx . t  ⇒ . Khi ó ta có:
dv = sin x dx  v = −cosx
 
u = x
 du = dx

A1 = x sin x − 3  − x cos x + 2 x cos x dx  . ∫
3 2
t  ⇒ .
  dv = cos x dx  v = sin x
 
3 2
( ∫ ) 3 2
A1 = x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x ) + c
Cách làm nhanh: Bi n i v d ng ∫ P ( x ) L ( x ) dx = ∫ P ( x ) du
A1 = x3 cos x dx = x 3 d ( sin x ) = x 3 sin x − sin x d ( x3 ) = x3 sin x − 3 x 2 sin x dx
∫ ∫ ∫ ∫
= x sin x + 3 x d ( cos x ) = x sin x + 3  x cos x − cos x d ( x ) 
∫ ∫
3 2 3 2 2
 
∫ ∫
3 2 3 2
= x sin x + 3x cos x − 6 x cos x dx = x sin x + 3x cos x − 6 x d ( sin x )
3 2
( ∫ ) 3 2
= x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x) + c
1 3 ( 5 x −1 ) 1  3 5 x −1
∫
• A2 = x 3 e 5x − 1 dx = ∫
x d e = xe − e5 x −1 d ( x3 ) 
∫
5 5 
1 1 3 2 ( 5x −1 ) 
=  x 3 e5x −1 − 3 x 2 e5x −1 dx  =  x 3 e5x −1 −
∫ x d e  ∫
5   5 5 
1 3 1 3 6
= x 3 e5x −1 −  x 2 e5x −1 − e5x −1 d ( x 2 )  = x 3 e5x −1 − x 2 e5x −1 +
∫ xe5x −1 dx ∫
5 25   5 25 25
1 3 6 1 3
= x 3 e5x −1 − x 2 e5x −1 +
5 25 125 ∫
x d ( e5x −1 ) = x 3 e5x −1 − x 2 e5x −1 +
5 25
6  5x −1 1 3 6 6 5x −1
+ xe − e5x −1 dx  = x 3 e5x −1 − x 2 e5x −1 +
∫ xe5x −1 − e +c
125   5 25 125 625
Nh n xét: N u P(x) có b c n thì ph i n l n s d ng tích phân t ng ph n.
211

3. Chương II: Nguyên hàm và tích phân − Tr n Phương
π2 /4 x 0 π2/4
• A3 = ∫ x sin x dx . t t = x ⇒t = x ⇒ 2
t 0 π/2
0
dx 2tdt
π2 π2 π2 π2
π2
∫ ∫ + 2 ∫ cos td ( t ) = 6 ∫ t cos t dt
3 3 3 3 2
A3 = 2 t sin t dt = −2 t d ( cos t ) = −2t cos t 0
0 0 0 0
π2 π2 π2 π2
π2 3π2 3π2
= 6 t 2 d ( sin t ) = 6t 2 sin t 0 − 6 sin td ( t 2 ) =
∫ ∫ ∫
−12 t sin t dt = + 12 td ( cos t ) ∫
0 0
2 0
2 0
π2
3π 2 π2 3π2 π2 3π2
=
2
+ 12t cos t 0 − 12 ∫
0
cos t dt =
2
− 12 sin t 0 =
2
− 12
π6 π 6 π 6 π 6
1 x cos3 x 1
∫ ∫ x d ( cos3 x ) = − ∫ cos
2
• A4 = x sin x cos xdx = − + 3
x dx
0
3 0
3 0 3 0
π6 π6
(1 − sin x ) d ( sin x ) = − π 3 + 1  sin x − sin x  = 11π − π 3

3
π 3 1
∫
2
=− + 
48 3 0
48 3  3 0 72 48
1
u = x 2 e x du = x ( x + 2 ) e x dx
2 x
x e dx  
• A5 = ∫ ( x + 2)
0
2
. t 
dv =
dx ⇒ 
v = −
1
 ( x + 2 )2  x+2
2 x 1 1 1 1
x e e e
+ xe dx = − + xe dx = − + xd ( e )
∫ ∫ ∫
x x x
A5 = −
x+20 0 3 0 3 0
1 1 1
e e e
∫
x x x
= − + xe 0
− e dx = − + e − e 0
=1−
3 0
3 3
2. D ng 2:
∫ P ( x ) {arcsin u; arccos u; arctg u; arc cotg u ; ln u ; log m u u = ax + b} dx
e
1
e
 e e

2
( ln x ) 2 d ( x 3 ) = 1  x3 ( ln x )2 1 − x3 d ( ln x ) 2 
∫
• B1 = x 2 ( ln x ) dx =
1
31 ∫ 3 1 

∫
1 dx  1  3  1 3 2 e 
e e
=  e3 − 2x 3 ln x ∫
2
 =  e − 2x ln x dx  =  e − ln x d ( x 3 )  ∫ ∫
3
 1
x  3
  1
 3
  31 

e 2 ( 3  3
3 e e e 3 e 3
= −  x ln x ) 1 − x3d ( ln x ) = e − 2 e3 + 2 x2 dx = e + 2 x3 = 5e − 2
∫ ∫
3 9 1  3 9
 91 9 27 1 27 27
212

4. Bài 8. Phương pháp tích phân t ng ph n
 1 + x  ( 2 ) 1  2 1 + x  1 + x 
12 12 12 12
 1+ x  1 2 
• B2 = ∫
0
x ln   dx =
 1− x  2 ∫0
ln 
 1− x 
 d x =  x ln
2 

 − x d  ln
1− x  0 0

 1 − x 
∫
12 12 2
1 1 − x dx 1  x 
∫ ∫
2
= ln 3 − x ⋅ ⋅ = ln 3 −   dx
8 0
1+ x 1− x 2
8 0
1+ x 
12 2 12
1  1  1  1 2 
= ln 3 −
8 ∫
0
1 −  dx = ln 3 −
 1+ x  8 ∫ 1 + (1 + x )
0 
2
−
1+ x 

dx
12
1  1  ln 3 3 5
= ln 3 −  x − − 2 ln 1 + x  = + 2 ln −
8  1+ x 0 8 2 6
1 1 1
• B3 = ∫ ln ( x + 1+ x 2 ) dx =  x ln ( x +
 1+ x 2 ) 0 − ∫ xd ln ( x +
  1 + x2 )

0 0
1 1
 x  dx x dx
= ln (1 + 2 ) − x  1 + ∫ = ln (1 + 2 ) − ∫
2  2 2
0  1+ x  x + 1+ x 0 1+ x
1 d (1 + x 2 )
1 1
= ln (1 + 2 ) − ∫ = ln (1 + 2 ) − 1 + x = ln (1 + 2 ) + 2 − 1
2
0
2 0 1+ x 2
x ln ( x + 1 + x 2 ) dx = 1 ln ( x +
1
• B4 = ∫ ∫ 1 + x2 ) d ( 1 + x2 )
2
0 1+ x 0
1 1
ln ( x + ) ∫ 1 + x d  ln x + 1 + x
( 
)
2 2 2 2
= 1+ x 1+ x − 
0
0
 
1
dx
= 2 ln (1 + 2 ) − ∫ 1 + x 1 + x 
2
 2  2
0  1+ x  x + 1+ x
1
= 2 ln (1 + 2 ) − dx = 2 ln (1 + 2 ) − 1
∫
0
u = ln x + 1 + x 2
( )
x ln ( x + 1 + x 2 ) dx .
1

• B5 = ∫
0 x + 1 + x2
t 
dv =
x dx =x ( 1 + x − x dx
2
)
2
 x + 1+ x
 x  (x + ) dx
⇒ du =  1 +  dx 1 + x2 =
2 2
 1+ x  1+ x
213

5. Chương II: Nguyên hàm và tích phân − Tr n Phương
v=
1
∫ (1 + x 2 )1 2 d (1 + x 2 ) − ∫ x 2 dx = 1 (1 + x 2 )3 2 − x 3 
 
2 3
1 1
1 2 
B5 =  (1 + x ) − x  ln x + 1 + x  −
2 32 1 ( 2 32 3 dx ( )
 1+ x ) − x  ∫
3
 
3 0 3 0 1+ x
2
(2 2 − 1) ln (1 + 2 ) 1 dx 1 x 3 dx
1 1
=
3
− +
3 0 1 + x2 3 0 1 + x2 ∫ ∫
(2 2 − 1) ln (1 + 2 ) 1 1 (1 + x ) − 1 (
1 1 2
d 1+ x ) ∫
2
= − arctg x +
3 3 0 60 1+ x 2
(2 2 − 1) ln (1 + 2 ) π 1 ( 2 12 2 −1 2 
1
 1 + x ) − (1 + x )  d (1 + x ) ∫
2
= − +
3 12 6 0
(2 2 − 1) ln (1 + 2 ) π 1  2 ( 2 32 2 12
1
= − +  1 + x ) − 2 (1 + x ) 
3 12 6  3 0
(2 2 − 1) ln (1 + 2 ) π 2 − 2
= − +
3 12 9
1 1
• B6 = ∫ x ln ( x + 1+ x 2 ) dx = 1 ∫ ln ( x + 1 + x2 ) d ( x2 )
0
2 0
1
 x 2 ln x + 1 + x 2 ( ) 1 2 
1
=  x d  ln x + 1 + x 
∫ ( )
2
− 

 2 0 2
 0
1 1 2
1 1 2 x  dx 1 1 x dx
= ln (1 + 2 ) − x 1+ = ln (1 + 2 ) −
∫ ∫
2 20 
2 
1+ x  x + 1+ x
2 2 2 0 1+ x
2
1 x 0 1
x 2 dx
Xét I = ∫
0 1 + x2
. t x = tg t ; t ∈  0, π ⇒
 2
 ) t 0 π/4
dx dt cos 2 t
1 2 π4 2 π4 2 π4 2
x dx tg t dt sin t sin t
⇒I= ∫
0 1+ x
2
= ∫
0 1 + tg t cos t
2
⋅ 2
= ∫ cos
0
3
t
dt = ∫ cos
0
4
t
d ( sin t )
π4 2 2 2 2 2 2 2
sin t d ( sin t ) u du 1  (1 + u ) − (1 − u ) 
= ∫
0 (1 − sin 2 t )2
= ∫
0 (1 − u 2 )2
=
4 ∫
0
 (1 + u ) (1 − u )  du
 
214

6. Bài 8. Phương pháp tích phân t ng ph n
2 2 2 2 2
1  1 1  1  1 1 2 
=
4 ∫0
 −  du =
1− u 1+ u  4 ∫ 0
(
 1 − u)
2
+ −
(1 + u ) 1 − u 2 
2

du
2 2
1 1 1 1+ u  2
=  − − 2 ln  = − ln (1 + 2 )
4 1− u 1+ u 1− u  0 2
1 1 1 1 2  2
⇒ B6 = ln (1 + 2 ) − I = ln (1 + 2 ) −  − ln (1 + 2 )  = − + ln (1 + 2 )
2 2 2 2 2  4
0 0 0
1 1 0 1
• B7 = ∫ x ln 1 − xdx = ∫ ln 1 − x d ( x2 ) = x2 ln 1 − x −8 − x 2 d ( ln 1 − x ) ∫
−8
2 −8 2 2 −8
0 0 2
1 −1 dx 1 x dx
= −32ln 3 −
2 −8 ∫
x2 ⋅ ⋅
2 1− x 1− x
= −32 ln 3 +
4 −8 1 − x ∫
1 1 − (1 − x )
0 2 0
1  1 
= −32ln 3 + ∫
4 −8 1 − x
dx = −32 ln 3 + ∫
4 −8 1 − x

− (1 + x )  dx

0
1 1 2 l 63
= −32ln 3 +  − ln 1 − x − x − 2 x  = −32 ln 3 + 6 + 2 ln 3 = 6 − 2 ln 3
4  −8
0 x −3 0
ln 1 − x
• B8 = ∫ (1 − x )
−3 1− x
dx . t t = 1− x ⇒ t 2 1
dx −2tdt
1 2 2
Khi ó ta có: B8 = ∫
2
ln t
t
3
( −2t dt ) = 2 ln t dt = 2 ln t d −1
∫ 2 ∫ 1 t 1
(t)
2 2 2 2
−2 ln t −1 dt 2
=
t 1
−2
1
t ∫
d ( ln t ) = − ln 2 + 2 2 = − ln 2 −
1 t
t1
= 1 − ln 2 ∫
3
1 ln x d ( x 2 + 1) 1
3 3
x ln x dx  −1 
• B9 = ∫ (x
1
2
+ 1)
2
=
2 1 ( x 2 + 1)∫2
=
21
ln x d  2 
 x + 1 ∫
3 3 3
− ln x 1 1 − ln 3 1 dx
=
2 ( x + 1) 1
2
+ 2
2 1 x +1 ∫
d ( ln x ) =
20
+
2 1 x ( x 2 + 1) ∫
− ln 3 1 ( x + 1) − x
3 2 2 3
− ln 3 1  1 x 
=
20
+ ∫
2 1 x ( x + 1)
2
dx =
20
+  − 2  dx
2 1  x x +1 ∫
3
− ln 3 1  1 2  9 ln 3
= + ln x − ( x + 1)  = −2
20 2 2 1 20
215

7. Chương II: Nguyên hàm và tích phân − Tr n Phương
3. D ng 3: Tích phân t ng ph n luân h i
1 1 1 3
∫
• C 1 = x 2 sin ( ln x ) dx =
3 ∫
sin ( ln x ) d ( x3 ) = x 3 sin ( ln x ) −
3 3
x d ( sin ( ln x ) ) ∫
1 1 dx 1 1
3 3 ∫
= x3 sin ( ln x ) − x3 cos ( ln x ) = x3 sin ( ln x ) − x2 cos ( ln x ) dx
x 3 3 ∫
1 3 1 1 3 1 3 1 3
= x sin ( ln x ) − cos ( ln x ) d ( x ) = x sin ( ln x ) − x cos ( ln x ) + x d ( cos ( ln x ) )
∫ ∫
3
3 9 3 9 9
1 3 1 3 1 2 1 3 1 3 1
3 9 9 3 ∫
= x sin ( ln x ) − x cos ( ln x ) − x sin ( ln x) dx = x sin ( ln x ) − x cos ( ln x ) − C1
9 9
10 1 1 1
⇒ C1 = x3 sin ( ln x ) − x3 cos ( ln x ) ⇒ C1 = 3x3 sin ( ln x ) − x3 cos ( ln x )  + c
 
9 3 9 10
π π π π
1 2x e2 x 1 2x e2π −1 1
∫ ∫ ∫
2x 2
• C2 = e sin x dx = e (1 − cos 2x ) dx = − e cos 2 x dx = − J
0
20 4 0 20 4 2
π π π π
1 2x 1 2x 1
∫ ∫ sin 2x d ( e ) ∫
2x
J = e 2 x cos 2 x dx = e d ( sin 2x ) = e sin 2x −
0
20 2 0 20
π π π π
1 2x 1 1
∫
= − e 2x sin 2x dx = ∫
e d ( cos 2x ) = e 2x cos 2x − cos 2x d ( e 2x ) ∫
0
20 2 0 20
2π π 2π 2π 2π
e −1 e −1 e −1 e −1
∫
2x
= − e cos 2x dx = − J ⇒ 2J = ⇒J=
2 0
2 2 4
e2 π − 1 1 e2π − 1 e2 π − 1 e2 π − 1
⇒ C2 = − J= − =
4 2 4 8 8
eπ eπ eπ
eπ
∫ cos ( ln x ) dx = ∫ xd ( cos ( ln x )) = − ( e + 1) + ∫ sin ( ln x ) dx
π
• C3 = x cos ( ln x ) 1 −
1 1 1
π π
e e
eπ
= − ( e + 1) + ∫ sin ( ln x ) dx = − ( e + 1) + ∫
− xd ( sin ( ln x ) )
π π
x sin ( ln x ) 1
1 1
π
e
−1 ( π )
= − ( e + 1) − cos ( ln x ) dx = − ( e + 1) − C3 ⇒ 2C3 = − ( e + 1) ⇒ C3 =
∫
π π π
e +1
1
2
eπ eπ eπ eπ
1 1 1 eπ −1 1
• C4 = ∫ cos ( ln x ) dx = ∫ [1 + cos ( 2ln x)] dx = x
2
− ∫
cos ( 2ln x) dx = − I
1
2 1
2 1 21 2 2
216

8. Bài 8. Phương pháp tích phân t ng ph n
eπ eπ eπ
eπ 2sin ( 2lnx)
∫
Xét I = cos ( 2 ln x ) dx = xcos( 2lnx) 1 − xd( cos( 2lnx) ) = e −1+ x ∫ ∫
π
dx
1 1 1
x
π π
e π e
e
= e − 1 + 2 ∫ sin ( 2 ln x ) dx = e − 1 + 2x sin ( 2 ln x ) 1 − 2 ∫ xd ( sin ( 2 ln x ) )
π π
1 1
π π
e e
2 cos ( 2 ln x )
∫ ∫
π π π
= e −1− 2 x dx = e − 1 − 4 cos ( 2 ln x ) dx = e − 1 − 4I
1
x 1
eπ − 1 eπ − 1 6 ( π
⇒ 5I = eπ − 1 ⇒ I = ⇒ C4 = e π − 1 + I = e π − 1 + = e − 1)
5 5 5
1 + sin x
• C5 = e x ∫ 1 + cos x
dx =
1 + sin x ( x ) x 1 + sin x
1 + cos x
d e =e ∫
1 + cos x
− e x d 1 + sin x
1 + cos x ∫ ( )
x x
1 + sin x x 1 + cos x + sin x x 1 + sin x e dx e sin x dx
∫ ∫ ∫
x
=e − e 2
dx = e − −
1 + cos x (1 + cos x ) 1 + cos x 1 + cos x (1 + cos x )2
x x
1 + sin x e dx e sin x dx
∫ ∫
x
=e − I − J (1) ; I = ;J=
1 + cos x 1 + cos x (1 + cos x ) 2
u = e x du = e x dx
e x sin x dx  
Xét J = ∫ (1 + cos x ) . t  sin x dx ⇒  −d (1 + cos x ) 1
∫
2
dv = 2 v = =
 (1 + cos x )  (1 + cos x ) 2
1 + cos x
x x x
e e dx e
⇒ J=
1 + cos x
− = ∫
1 + cos x 1 + cos x
−I (2). Thay (2) vào (1) ta có:
1 + sin x  ex  x 1 + sin x e
x
⇒ C5 = e x −I− − I + c = e − +c
1 + cos x  1 + cos x  1 + cos x 1 + cos x
π π π
sin 2 x
π
1 −x 1 −x 1 −x
• C6 = ∫
0
e x
dx =
20 ∫
e (1 − cos 2 x ) dx =
20
e dx −
20
e cos 2 x dx ∫ ∫
−x π π −π π −π
−e 1 −x 1− e 1 −x 1− e 1
=
2 0
− ∫
20
e cos 2 x dx =
2
−
20 ∫
e cos 2 x dx =
2
−
2
J
π π −x π π
1 −x e sin 2x 1
∫ ∫ sin 2x d ( e )
∫
−x −x
J= e cos 2 x dx = e d ( sin 2x ) = −
0
20 2 0 20
π π π π
1 −x −1 − x e− x cos 2 x 1
=
20 ∫
e sin 2 x dx =
4 0
e d ( cos 2 x ) = −
4 0
+∫40
cos 2 x d ( e − x ) ∫
217

9. Chương II: Nguyên hàm và tích phân − Tr n Phương
π
1 − e −π 1 − x 1 − e −π 1 5 1 − e −π 1 − e −π
=
4
−
40 ∫
e cos 2 x dx =
4
− J ⇒ J=
4 4 4
⇒J=
5
1 − e −π 1 1 − e −π 1 − e −π 2 (
⇒ C6 = − J= − = 1 − e −π )
2 2 2 10 5
a
• C7 = ∫
0
a 2 − x 2 dx ; ( a > 0 )
a2 − ( a2 − x2 )
a a a
a
x 2 dx
C7 = x a 2 − x 2 0 − x d ( a2 − x2 ) =
∫ ∫ = ∫ dx
0 0 a2 − x2 0 a2 − x2
a a a a
dx x πa 2
= a2 ∫
0 a2 − x2
− ∫
0
a 2 − x 2 dx = a 2 arcsin
a 0
− ∫
0
a 2 − x 2 dx =
2
− C7
2 2
πa πa
⇒ 2C7 = ⇒ C7 =
2 4
a
• C8 = ∫
0
a 2 + x 2 dx ; ( a > 0 )
a a
a
( ) = a2 x2
0 − ∫ xd ∫
2 2 2 2
C8 = x a + x a +x 2− dx
0 0 a2 + x2
a
( a2 + x2 ) − a2 a a
dx
∫ ∫ ∫
2 2 2 2 2
=a 2− dx = a 2− a + x dx + a
2 2
0 a +x 0 0 a + x2
2
a a
= a 2 2 + a 2 ln x + a 2 + x 2 0
− ∫ a 2 + x 2 dx = a 2 2 + a 2 ln (1 + 2 ) − C8
0
2 + ln (1 + 2 ) 2
⇒ 2C8 = a 2 2 + a 2 ln (1 + 2 ) ⇒ C8 = a
2
a u = x du = dx
 
∫
2 2 2
• C9 = x a + x dx ; ( a > 0 ) . t  ⇒ 3
dv = x a 2 + x 2 dx  v = 1 ( a + x ) 2
2 2
0   3
3 a a3
x 2 1 ( 2
C9 = ( a + x ) 2 a + x ) 2 dx
∫
2 2
−
3 0 30
a a
2 2 4 a2 1 2 2 2 2 4 a2 1
∫ ∫
2 2 2
= a − a + x dx − x a + x dx = a − C8 − C9
3 3 0
30 3 3 3
218

10. Bài 8. Phương pháp tích phân t ng ph n
2 + ln (1+ 2 ) 3 2 − ln (1+ 2 ) 3 2 − ln (1+ 2 )
2
4 2 2 4 a
⇒ C13 = a − ⋅ = ⇒ C9 =
3 3 3 2 6 8
a u = x du = dx
 
∫
2 2 2
• C 10 = x a − x dx ; ( a > 0 ) . t  ⇒ 3
dv = x a 2 − x 2 dx  v = −1 ( a − x ) 2
2 2
0   3
a
1 
3 a 3 a a
−x ( 2 1 ( 2
C10 = a − x2 ) 2 + a − x 2 ) 2 dx =  a 2 a 2 − x 2 dx + x 2 a 2 − x 2 dx 
∫ ∫ ∫
3 0 30 30
 0


2 2 4 4
a 1 2 a πa πa
= C7 + C10 ⇒ C10 = C7 = ⇒ C10 =
3 3 3 3 12 8
2a 2a 2a
• C 11 = ∫ x 2 − a 2 dx = x x 2 − a 2 a 2 − ∫ x d ( x2 − a2 )
a 2 a 2
a + (x − a )
2a 2a 2 2 2
x
= (2 3 − 2 ) a − ∫ dx = ( 2 3 − 2 ) a − ∫
2 2
x dx
2 2 2 2
a 2 x −a a 2 x −a
2a 2a
dx
= (2 3 − 2 ) a − a ∫ ∫
2 2 2 2
− x − a dx
2 2
a 2 x −a a 2
2a 2a
= ( 2 3 − 2 ) a − a ln x + x − a ∫
2 2 2 2 2 2
a 2
− x − a dx
a 2
2+ 3
= ( 2 3 − 2 ) a − a ln
2 2
− C11
1+ 2
a  2+ 3
2
2+ 3
⇒ 2C11 = ( 2 3 − 2 ) a − a ln ( 2 3 − 2 ) − ln
2 2
⇒ C11 = 
1+ 2 2  1+ 2 
π 2 π 2 π 2 π 2
dx
∫ cotg x d ( sin x )
1 cotg x 1
• C 12 = ∫
π 4
3
sin x
=−
π 4
∫
sin x
d ( cotg x ) = −
sin x π 4
+
π 4
π2 π2
cos x 1
∫ sin x  sin 
1
=− 2 − ∫ cotg x sin
π4
2
x
dx = − 2 −
π4


2
x
− 1 dx

π2 π2 π2
dx dx sin x dx
=− 2 + ∫
π4
−
sin x π 4 sin 3 x ∫
=− 2 +
π4
2
1 − cos x
− C12 ∫
π2
1 1 + cos x − 2 + ln (1 + 2 )
⇒ 2C12 = − 2 − ln = − 2 + ln (1 + 2 ) ⇒ C12 =
2 1 − cos x π4 2
219

11. Chương II: Nguyên hàm và tích phân − Tr n Phương
4. D ng 4: Các bài toán t ng h p
3
x3 ( x 2 + 2 ) x 3 ( x 2 + 1)
3 3 3
x 5 + 2x 3 x3
• D1 = ∫
0 x2 + 1
dx = ∫
0 x2 + 1
dx = ∫
0 x2 + 1
dx + ∫
0 x2 + 1
dx
3 3
x dx
∫ ∫x
2 2 2
= x .x x + 1 dx + =I+J
2
0 0 x +1
3 u = x 2 du = 2x dx
 
∫x ⇒
2 2
Xét I = .x x + 1 dx . t  1 2 32
0 dv = x x 2 + 1 dx  v = ( x + 1)
  3
3 3 3
1 2 2 32 2 32 1 32
I = x ( x + 1) ∫ x ( x + 1) dx = 8 − ∫ ( x + 1) d ( x + 1)
2 2 2
−
3 0 3 0
3 0
3
2( 2 2
( 32 − 1) = 58
52
=8− x + 1) =8−
15 0 15 15
u = x 2
3
x dx 
 du = 2x dx

∫x x dx ⇒ 
2
Xét J = . t 
0 x2 + 1 dv = v = x 2 + 1


 x2 + 1
3 3 3 3
2 2 32 4
∫ 2x ∫ + 1 d ( x + 1) = 6 − ( x + 1)
2 2 2 2 2
J=x x +1 0 − x + 1 dx = 6 − x =
0 0
3 0 3
58 4 26
⇒ D1 = I + J = + =
15 3 5
2
1 d ( 1 + x3 )
2 2 2
1 + x3 1  = − 1+ x
3
• D2 = ∫ 4
dx = − 1 + x3 d  13
 ∫  + ∫
1
x 31 x  3 x3 1 31 x3
2 1 1 d (1 + x )
2 2 2 3
2 1 1 x dx
= − +
3 8 2 1 x3 1 + x3
=
3
− + ∫
8 6 1 x3 1 + x3 ∫
d (u2 )
3 3
2 1 1 2 1 1 du
= − +
3 8 6 ∫ (u 2
− 1) u
=
3
− +
8 3 ∫u 2
−1
2 2
3
2 1 1 u −1 2 1 1 1 
= − + ln = − +  ln + 2 ln (1 + 2 ) 
3 8 3 u +1 2 3 8 3 2 
220

12. Bài 8. Phương pháp tích phân t ng ph n
π 2 π 2
sin 2 x 1
∫e ∫ 2 cos
2
3
• D3 = sin x cos x dx = 2
x ( 2 sin x cos x ) esin x
dx
0
4 0
π2 π2 π2
1 ) = 1 (1 + cos 2x ) esin 1
(1 + cos 2x ) d ( esin
2 2
sin 2 x
∫ ∫e
x x
= − d (1 + cos 2x )
4 0
4 0 4 0
π2 π2 π2
−1 1 1 1 1 1 sin e
∫ d (e ) = − + e
sin 2 x 2 2
∫e
sin x x
= + sin 2x dx = − + = −1
2 2 0
2 2 0
2 2 0 2
π 2 π 2
• D4 = ∫
π 3
cos x ln ( 1 − cos x ) dx =
π
∫ ln (1 − cos x ) d ( sin x )
3
π2 π2
π2 3 sin x dx
= sin x ln (1 − cos x )
π3
− ∫
π3
sin xd ( ln (1 − cos x ) ) =
2
ln 2 − sin x
π3
∫
1 − cos x
π2 2 π2
3 1 − cos x 3
=
2
ln 2 −
π3
1 − cos x ∫
dx =
2
ln 2 − (1 + cos x ) dx
π3
∫
3 π2 3 π 3
= ln 2 − ( x + sin x ) = ln 2 − − 1 +
2 π3 2 6 2
π3 π3 π3
π3
• D5 =
π4
∫ sin x ln( tg x) dx = −π∫ ln ( tg x) d ( cos x) = − cos x ln ( tg x) π
4
4
+
π4
∫ cos x d ( ln ( tg x))
π3 π3 π3
1 cos x dx 1 dx 1 sin x dx
= − ln 3 +
4 2
π 4 cos x tg x
∫
= − ln 3 +
4 π4
sin x
= − ln 3 +
4 2
π 4 sin x
∫ ∫
π3 π3
1 d ( cos x ) 1 1 1 + cos x 3
= − ln 3 − 2 ∫ = − ln 3 − ln = ln (1 + 2 ) − ln 3
4 π 4 1 − cos x
4 2 1 − cos x π4 4
π4 π 4 π 4 π 4 π 4
x + sin x d (1 + cos x )
∫ xd ( tg 2 )
sin x dx x dx x
• D6 = ∫
0
1 + cos x
dx = ∫
0
1 + cos x
+ ∫ 2 cos
0
2 x
=− ∫0
1 + cos x
+
0
2
π4 π4 π4
 x x 4 π π x
=  − ln 1 + cos x + x tg  −
 20 ∫
0
tg
2
dx = ln + tg + 2 ln cos
2+ 2 4 8 2 0
π
= ln
4
+ ( 2 − 1) + ln 2 + 2 = π ( 2 − 1) + ln1 = π ( 2 − 1)
2+ 2 4 4 4 4
221