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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
MT S K THUT "PHN TCH " M TA THNG GP KHI I TM
NGUYN HM HOC TNH TCH PHN.
Th d 1 : Tm nguyn hm A1 =
dx
(x+ 1) (x+ 2)
Ta gi s rng :
1
(x+ 1) (x+ 2)=
x+ 1+
x+ 2Ta i tm 2 h s , theo ba cch nh sau :
Cch 1 : = limx1
x+ 1
(x+ 1) (x+ 2)= lim
x11
x+ 2=
1
1 + 2 = 1
= limx2
x+ 2
(x+ 1) (x+ 2)= lim
x21
x+ 1=
1
2 + 1 = 1Cch 2 :
Cho x = 0 ta c :1
(0 + 1) (0 + 2)=
0 + 1+
0 + 2 1
2= +
1
2
x = 1 ta c :1
(1 + 1) (1 + 2)=
1 + 1+
1 + 2 1
6=
1
2 +
1
3
Do m ta suy ra 2 h s , bng cch i gii h :
+1
2 =
1
21
2 +
1
3 =
1
6
{
= 1
= 1Cch 3 :
Ta gi s rng :
1
(x+ 1) (x+ 2)=
x+ 1+
x+ 2
1(x+ 1) (x+ 2)
= (x+ 2) + (x+ 1)
(x+ 1) (x+ 2)=x ( + ) + 2 +
(x+ 1) (x+ 2)
Cn bng cc h s 2 v m ta c h :
{ + = 0
2 + = 1{
= 1
= 1Do m ta suy ra :
1
(x+ 1) (x+ 2)=
1
x+ 1 1x+ 2
Vy :
dx
(x+ 1) (x+ 2)=
dx
x+ 1
dx
x+ 2= ln |x+ 1| ln |x+ 2|+ c = ln
x+ 1x+ 2
+ cTh d 2 : Tm nguyn hm A2 =
x+ 2
x (x 2) (x+ 5)dx
Ta gi s rng :
x+ 2
x (x 2) (x+ 5) =
x+
x 2 +
x+ 5Cch 1 :
= limx0
x (x+ 2)
x (x 2) (x+ 5) = limx0x+ 2
(x 2) (x+ 5) =0 + 2
(0 2) (0 + 5) = 1
5
= limx2
(x 2) (x+ 2)x (x 2) (x+ 5) = limx2
x+ 2
x (x+ 5)=
2 + 2
2 (2 + 5)=
2
7
= limx5
(x+ 5) (x+ 2)
x (x 2) (x+ 5) = limx5x+ 2
x (x 2) =5 + 2
5 (5 2) = 3
35Cch 2 :
Cho x = 1 ta c : 1 + 21 (1 2) (1 + 5) =
1 +
1 2 +
1 + 5 1
3 +
1
4 =
1
12
x = 1 ta c :1 + 2
1 (1 2) (1 + 5) =
1+
1 2 +
1 + 5 + 1
6 = 1
2
1
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
x = 3 ta c :3 + 2
3 (3 2) (3 + 5) =
3+
3 2 +
3 + 5 1
3 + +
1
8 =
5
24 tm c cc h s , , ta gii h phng trnh :
13 +
1
4 =
1
12
+ 16 = 1
21
3 + +
1
8 =
5
24
= 1
5
=2
7
= 335Cch 3 :
Gi s rng :
x+ 2
x (x 2) (x+ 5) =
x+
x 2 +
x+ 5
x+ 2x (x 2) (x+ 5) =
(x 2) (x+ 5) + x (x+ 5) + x (x 2)x (x 2) (x+ 5)
x+ 2x (x 2) (x+ 5) =
x2 ( + + ) + x (3 + 5 2) 10x (x 2) (x+ 5)
Cn bng cc h s 2 v m ta c h :
+ + = 0
3 + 5 2 = 110 = 2
= 1
5
=2
7
= 335
Do :
x+ 2
x (x 2) (x+ 5) = 1
5x+
2
7 (x 2) 3
35 (x+ 5)
Vy :
x+ 2
x (x 2) (x+ 5)dx = 1
5
dx
x+
2
7
dx
x 2 3
35
dx
x+ 5
= 15
ln |x|+ 27
ln |x 2| 335
ln |x+ 5|+ c
Th d 3 : Tm nguyn hm A3 =
x2
(3x2 2x+ 5) (x+ 1) dxTa cn nh phng trnh bc 2 c dng : ax2 + bx+ c
c 2 nghim x1, , x2 th c biu din di
dng : ax2 + bx+ c = a (x x1) (x x2)Do ta vit : 3x2 2x+ 5 = 3 (x
1)
(x+
5
3
)Ta gi s rng :
x2
(3x2 2x+ 5) (x+ 1) = x2
3 (x 1)(x+
5
3
)(x+ 1)
=
x 1 +
x+5
3
+
x+ 1
= limx1
x2 (x 1)3 (x 1)
(x+
5
3
)(x+ 1)
= limx1
x23(x+
5
3
)(x+ 1)
= 116
= limx 5
3
x2(x+
5
3
)3 (x 1)
(x+
5
3
)(x+ 1)
= limx 5
3
[ x
2
3 (x 1) (x+ 1)
]= 25
48
= limx1
x2 (x+ 1)3 (x 1)
(x+
5
3
)(x+ 1)
= limx1
x23 (x 1)
(x+
5
3
) = 1
4
Cch 2 :
2
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Cho x = 0 ta c :x2
(3x2 2x+ 5) (x+ 1) =
x 1 +
x+5
3
+
x+ 1 + 3
5 + = 0
x = 2 ta c :x2
(3x2 2x+ 5) (x+ 1) =
x 1 +
x+5
3
+
x+ 1 4
33= +
3
11 +
1
3
x = 3 ta c :x2
(3x2 2x+ 5) (x+ 1) =
x 1 +
x+5
3
+
x+ 1 9
112=
1
2 +
3
14 +
1
4
t m ta c h :
+ 3
5 + = 0
+3
11 +
1
3 = 4
331
2 +
3
14 +
1
4 = 9
112
= 1
16
= 2548
=1
4Cch 3 :
Gi s rng :
x2
(3x2 2x+ 5) (x+ 1) = x2
3 (x 1)(x+
5
3
)(x+ 1)
=
x 1 +
x+5
3
+
x+ 1
x2
3 (x 1)(x+
5
3
)(x+ 1)
=(x+
5
3
)(x+ 1) + (x 1) (x+ 1) + (x 1)
(x+
5
3
)(x 1)
(x+
5
3
)(x+ 1)
x2 = x2 ( + + ) + x(
8
3 +
2
3)
+5
3 5
3
Cn bng cc h s 2 v ta c h :
+ + = 18
3 +
2
3 = 0
5
3 5
3 = 0
= 1
16
= 2548
=1
4Do m ta c :
x2
(3x2 2x+ 5) (x+ 1) = x2
3 (x 1)(x+
5
3
)(x+ 1)
= 116 (x 1)
25
48(x+
5
3
) + 14 (x+ 1)
Suy ra :
x2
(3x2 2x+ 5) (x+ 1) dx = 1
16
dx
x 1 25
48
dx
x+5
3
+1
4
dx
x+ 4
= 116
ln |x 1| 2548
lnx+ 5
3
+ 14
ln |x+ 4|+ c
Th d 4 : Tm nguyn hm A4 =
x 1
(x2 + 4x+ 5) (x2 4) dxTa cn ch rng phng trnh : ax2 + bx+ c = 0
vi = b2 4ac < 0 th ta vit :ax2 + bx+ c = a (x x1) (x x2) trong :
x1 = + i, x2 = i, i2 = 1Do m ta c : x2 + 4x+ 5 = (x+ 2 + i) (x+ 2
i)Ta gi s rng :
x 1(x2 + 4x+ 5) (x2 4) =
x 1(x+ 2 + i) (x+ 2 i) (x2 4) =
x+ 2 + i+
x+ 2 i +
x 2 +
x+ 2
= limx2i
(x 1) (x+ 2 + i)(x+ 2 + i) (x+ 2 i) (x2 4) = limx2i
x 1(x+ 2 i) (x2 4) =
13
34 1
34i
3
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
= limx2+i
(x 1) (x+ 2 i)(x+ 2 + i) (x+ 2 i) (x2 4) = limx2+i
x 1(x+ 2 + i) (x2 4) =
13
34+
1
34i
= limx2
(x 1) (x 2)(x2 + 4x+ 5) (x 2) (x+ 2) = limx2
x 1(x2 + 4x+ 5) (x+ 2)
=1
68
= limx2
(x 1) (x+ 2)(x2 + 4x+ 5) (x 2) (x+ 2) = limx2
x 1(x2 + 4x+ 5) (x 2) =
3
4
Do m ta c :
x 1(x2 + 4x+ 5) (x2 4) =
1334 1
34i
x+ 2 + i+13
34+
1
34i
x+ 2 i +1
68x 2 +
3
4x+ 2
=13x 27
17 (x2 + 4x+ 5)+
1
68 (x 2) +3
4 (x+ 2)Cch 2 :
Gi s rng :
x 1(x2 + 4x+ 5) (x2 4) =
x+
x2 + 4x+ 5+
x 2 +
x+ 2
Cho x = 0 ta c :1
5 1
2+
1
2 =
1
20
x = 1 ta c :1
10 +
1
10 + 1
3 = 0
x = 3 ta c :3
26 +
1
26 + +
1
5 =
1
65
x = 4 ta c :4
37 +
1
37 +
1
2+
1
6 =
1
148 t m ta c c h :
1
5 1
2+
1
2 =
1
201
10 +
1
10 + 1
3 = 0
3
26 +
1
26 + +
1
5 =
1
654
37 +
1
37 +
1
2+
1
6 =
1
148
= 25 + +
1
101
10 1
30 2
3 = 1
303
26 27
650 +
6
5 = 3
6504
37 22
555 +
2
3 = 11
1110
= 1317
= 2717
=1
68
=3
4
Do :
x 1(x2 + 4x+ 5) (x2 4) =
13x 2717 (x2 + 4x+ 5)
+
68 (x 2) +3
4 (x+ 2)Cch 3 :
Gi s :
x 1(x2 + 4x+ 5) (x2 4) =
x+
x2 + 4x+ 5+
x 2 +
x+ 2 x 1 = (x+ ) (x2 4)+ (x2 + 4x+ 5) (x+ 2) + (x2 + 4x+ 5) (x
2) x 1 = x3 ( + + ) + x2 ( + 6+ 2) + x (4 + 13 3) + (4 + 10 10)
Cn bng cc h s 2 v ta c h :
+ + = 0
+ 6+ 2 = 0
4 + 13 3 = 14 10+ 10 = 1
= + 6+ 2 ( ) = 04 + 13 3 ( ) = 14 10+ 10 ( ) = 1
= 2 + + 4 = 0 + 16 = 110 + 4 20 = 1
= 1317
= 2717
=1
68
=3
4
4
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Do :
x 1(x2 + 4x+ 5) (x2 4) =
13x 2717 (x2 + 4x+ 5)
+1
68 (x 2) +3
4 (x+ 2)
Vy :
x 1
(x2 + 4x+ 5) (x2 4) dx = 13
34
2x+ 5413
x2 + 4x+ 5dx+
1
68
dx
x 2 +3
4
dx
x+ 2
= 1334
(2x+ 4)
x2 + 4x+ 5dx 1
17
dx
x2 + 4x+ 5+
1
68
dx
x 2 +3
4
dx
x+ 2
= 1334
d(x2 + 4x+ 5
)x2 + 4x+ 5
117
dx
(x+ 2)2 + 1+
1
68
dx
x 2 +3
4
dx
x+ 2
= 1334
lnx2 + 4x+ 5 1
17arctan (x+ 2) +
1
68ln |x 2|+ 3
4ln |x+ 2|+ c
Th d 5 : Tm nguyn hm A5 =
x
x3 + 1dx
Ta gi s rng :
x
x3 + 1=
x
(x+ 1) (x2 x+ 1) =
x+ 1+
x 12
3
2i
+
x 12
+
3
2i
= limx1
x
x2 x+ 1 = 1
3
= limx 1
2+32i
x
(x+ 1)
(x 1
2+
3
2i
) = 16
3
6i
= limx 1
232i
x
(x+ 1)
(x 1
2
3
2i
) = 16
+
3
6i
Do :
x
x3 + 1= 1
3 (x+ 1)+
1
6
3
6i
x 12
3
2i
+
1
6+
3
6i
x 12
+
3
2i
= 13 (x+ 1)
+x+ 1
3 (x2 x+ 1)
Cch 2 :
Gi s :
x
x3 + 1=
x+ 1+
x+
x2 x+ 1Cho x = 0 ta c : 0 = +
x = 1 ta c :1
2=
1
2 + +
x = 2 ta c :2
9=
1
3 +
2
3 +
1
3
Do m ta c h :
+ = 01
2 + + =
1
21
3 +
2
3 +
1
3 =
2
9
= 1
3
=1
3
=1
3
Vy :
x
x3 + 1= 1
3 (x+ 1)+
x+ 1
3 (x2 x+ 1)Cch 3 :
Ta cng gi s rng :
x
x3 + 1=
x+ 1+
x+
x2 x+ 1 x = (x2 x+ 1)+ (x+ ) (x+ 1)5
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Cn bng cc h s m ta c h :
+ = 0
+ + = 1 + = 0
= 1
3
=1
3
=1
3
Vy :
x
x3 + 1dx = 1
3
dx
x+ 1+
1
6
2x+ 2
x2 x+ 1dx
= 13
dx
x+ 1+
1
6
2x+ 1
x2 x+ 1dx+1
6
dx
x2 x+ 1= 1
3
dx
x+ 1+
1
6
d(x2 x+ 1)x2 x+ 1 +
1
6
dx(
x 12
)2+
3
4
= 13
ln |x+ 1|+ 16
lnx2 x+ 1+ 1
3
3arctan
2x 13
+ c
Th d 6 : Tm nguyn hm A6 =
dx
x4 + 1Gi s rng : x4 + 1 =
(x2 + px+ 1
) (x2 px+ 1) = x4 + (2 p2)x+ 1ng nht 2 v ta c : 2 p2 = 0 p =
2
Do ta vit : x4 + 1 =(x2 +
2x+ 1
) (x2
2x+ 1
) 1(
x2 +
2x+ 1) (x2 2x+ 1) =
=
x+
2
2+
2
2i
+
x+
2
2
2
2i
+
x
2
2+
2
2i
+
x
2
2
2
2i
Tm cc h s : , , , nh sau :
= limx
2222i
1(x+
2
2
2
2i
)(x2 2x+ 1)
= 28 +
2
8i
= limx
22+22i
1(x+
2
2+
2
2i
)(x2 2x+ 1)
= 28
2
8i
= limx
2222i
1(x
2
2
2
2i
)(x2 +
2x+ 1
) = 28 +
2
8i
= limx
22+22i
1(x
2
2+
2
2i
)(x2 +
2x+ 1
) = 28
2
8i
Vy :
1
x4 + 1=
(2
8+
2
8i
)(x+
2
2
2
2i
)+
(2
8
2
8i
)(x+
2
2+
2
2i
)x2 +
2x+ 1
+
6
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
+
(
2
8+
2
8i
)(x
2
2
2
2i
)+
(
2
8
2
8i
)(x
2
2+
2
2i
)x2 2x+ 1
=
2
4x+
1
2x2 +
2x+ 1
+
2
4x+
1
2x2 2x+ 1Cch 2 : Ta gi s rng :
1
x4 + 1=
x+
x2 +
2x+ 1+
x+
x2 2x+ 1Cho x = 1 ta c : 1
2= 1
22 +1
22 1
2 +
2+
1
2 +
2
x = 0 ta c : + = 1
x = 1 ta c :1
2=
1
2 +
2 +
1
2 +
2 1
22+1
22
x =
2 ta c :1
5=
1
5
(2 +
)+(
2+ )
t ta c h :
+ = 1
122 +
1
22 1
2 +
2+
1
2 +
2 =
1
21
2 +
2 +
1
2 +
2 1
22+1
22 =1
21
5
(2 +
)+(
2+ )
=1
5
+ = 1
22 + 2+2 = 0 + +
(3 + 2
2)+
(3 + 2
2) =
2 +
2
22 + + 5
2+ 5 = 1
+ = 1
22 + 2 = 2a (2 + 22) + (3 + 22) = 4 + 32
22 4 + 52 = 4
=
2
4
= =1
2
=
2
4
Vy ta c c :
1
x4 + 1=
2
4x+
1
2x2 +
2x+ 1
+
2
4x+
1
2x2 2x+ 1Cch 3 :
Ta cng gi s rng :
1
x4 + 1=
x+
x2 +
2x+ 1+
x+
x2 2x+ 1 1 = (x+ ) (x2 2x+ 1)+ (x+ ) (x2 +2x+ 1) 1 = ( + )x3 +
(2 + +2+ )x2 + (2 + +2)x++ Cn bng cc h s 2 v m ta c c h sau :
+ = 0
2 + +2+ = 02 + +2 = 0 + = 1
+ = 0
2 +2 = 1 22 + = 2 = 1
=
2
4
= =1
2
=
2
4
7
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Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Do :
1
x4 + 1=
2
4x+
1
2x2 +
2x+ 1
+
2
4x+
1
2x2 2x+ 1
Vy ta c :
dx
x4 + 1=
12
2x+
1
2
x2 +
2x+ 1dx+
12
2x+
1
2
x2 2x+ 1 dx
Ta c :
12
2x+
1
2
x2 +
2x+ 1dx =
1
4
2
2x+
2
x2 +
2x+ 1dx+
1
4
dx
x2 +
2x+ 1
=1
4
2
d(x2 +
2x+ 1
)x2 +
2x+ 1
+1
4
dx(
x+12
)2+
1
2
=1
4
2ln(x2 +
2x+ 1
)+
1
2
2arctan
(2x+ 1
)+ c1
Li c :
12
2x+
1
2
x2 2x+ 1 dx =1
4
2
2x+2x2 2x+ 1 dx+
1
4
dx
x2 2x+ 1= 1
4
2
d(x2 2x+ 1)x2 +
2x+ 1
+1
4
dx(
x 12
)2+
1
2
= 14
2ln(x2
2x+ 1
)+
1
2
2arctan
(2x 1)+ c2Vy :
dx
x4 + 1=
1
4
2ln
(x2 +
2x+ 1
x2 2x+ 1
)+
1
2
2arctan
(2x+ 1
)+
1
2
2arctan
(2x 1)+ cTh d 7 : Tm nguyn hm A7 =
dx
x8 + 1Gi s rng : x8 + 1 =
(x4 + px2 + 1
) (x4 px2 + 1) = x8 + (2 p2)x4 + 1ng nht 2 v ta c : 2 p2 = 0 p
=
2
x8 + 1 = (x4 +2x2 + 1) (x4 2x2 + 1)Ta c : x4 +
2x2 + 1 =
(x2 + qx+ 1
) (x2 qx+ 1) = x4 + (2 q2)x2 + 1ng nht 2 v ta c : 2 q2 =
2 q =
2
2
x4 +
2x2 + 1 =(x2 +
2
2x+ 1)(
x2
2
2x+ 1)Ta c : x4
2x2 + 1 =
(x2 + rx+ 1
) (x2 rx+ 1) = x4 + (2 r2)x2 + 1ng nht 2 v ta c c : 2 r2 =
2 r =
2 +
2
x4
2x2 + 1 =(x2 +
2 +
2x+ 1)(
x2
2 +
2x+ 1)Do m ta c :
x8+1 =(x2 +
2
2x+ 1)(
x2
2
2x+ 1)(
x2 +
2 +
2x+ 1)(
x2
2 +
2x+ 1)
1x8 + 1
=1
8
22x+ 2
x2 +
22x+ 1+
1
8
22x+ 2x2
22x+ 1
+1
8
2 +
2x+ 2
x2 +
2 +
2x+ 1+
+1
8
2 +
2x+ 2
x2
2 +
2x+ 1
8
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Ta ch :
dx
ax2 + bx+ c=
1
a
dx
x2 +b
ax+
c
a
, 4ac b2 > 0
=1
a
dx(
x b2a
)2+
(4ac b2
2a
)2 = 1a
dx
(x p)2 + q2(1)
t : p =b2a, q =
4ac b2
2a, x = p+ qt
Do : (1) =1
aq
1
t2 + 1dt =
24ac b2
1
t2 + 1dt =
24ac b2 arctan t
=2
4ac b2 arctan2ax+ b4ac b2 + C
Trong ta t : t =x pq
=2ax+ b4ac b2Li c :
xdx
ax2 + bx+ c=
1
2a
2ax+ b bax2 + bx+ c
dx =1
2a
2ax+ b
ax2 + bx+ cdx b
2a
1
ax2 + bx+ cdx
=1
2alnax2 + bx+ c b
2a
dx
ax2 + bx+ c+ C
Vi : a = c = 1, = 4 b2 > 0 ta c :Ax+B
x2 + bx+ 1dx = A
xdx
x2 + bx+ 1+B
dx
x2 + bx+ 1
=2B Ab
4 b2 arctan2x+ b4 b2 +
A
2lnx2 + bx+ 1+ CDo : A7.1 =
1
8
22x+ 2
x2 +
22x+ 1dx =
=
2 +
2
8arctan
2x+
222 +
2+
2216
ln(x2 +
2
2x+ 1)
+ C1
A7.2 =1
8
22x+ 2x2
22x+ 1
dx =
=
2 +
2
8arctan
2x
222 +
2
2216
ln(x2
2
2x+ 1)
+ C2
A7.3 =
2 +
2x+ 2
x2 +
2 +
2x+ 1dx =
=
22
8arctan
2x+
2222
+
2 +
2
16ln(x2 +
2 +
2x+ 1)
+ C3
A7.4 =
2 +2x+ 2x2
2 +
2x+ 1dx =
=
22
8arctan
2x
2222
2 +
2
16ln(x2
2 +
2x+ 1)
+ C4
Do m ta c : A7 = A7.1 + A7.2 + A7.3 + A7.4
Cch 2 :
Ta c : x8 + 1 = 0 x8 = 1 = cos (pi + k2pi) + i sin (pi +
k2pi)
9
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
x = cos(pi + k2pi
8
)+ i sin
(pi + k2pi
8
), vi : k = 0, .., 7
Ta c : sinpi
8= sin
7pi
8= cos
3pi
8=
1
2
2
2, sin3pi
8= sin
5pi
8= cos
pi
8=
1
2
2 +
2 ,
cos5pi
8= cos
7pi
8= 1
2
2
2 ,
Do m ta c :
dx
1 + x8= 1
8
3k=0
(ln
(x2 2x cos
((2k + 1) pi
8
)+ 1
))cos
((2k + 1) pi
8
)+
+1
4
3k=0
arctan x sin (2k + 1) pi8
1 x cos (2k + 1) pi8
sin((2k + 1) pi
8
)+ C, k = 0, ..., 7
Qua cc th d trn ta nhn thy cc h s : , , , ..., l cc h s "bt nh"
tc l "nhng h
s khng h thay i " d ta c thay i "gi tr ca bin x" . tm hiu r vn
ny ta s
lm quen dng ton ny vi cc bi tp mu tng t nh sau :
Bi 1 : Tm nguyn hm I1 =
x+ 27
(x 3) (x+ 3) dx
Gi s rng :
x+ 27
(x 3) (x+ 3) =
x 3 +
x+ 3 x+ 27 = (x+ 3) + (x 3) = x ( + ) + (3 3)
Cn bng h s 2 v ta c h :
{ + = 1
= 9 {
= 5
= 4Vy :
x+ 27
(x 3) (x+ 3) dx =
5
x 3 dx
4
x+ 3dx = 5 ln |x 3| 4 ln |x+ 3|+ c
Bi 2 : Tm nguyn hm I2 =
8x 17
(x 3) (x+ 4) dx
Gi s rng :
8x 17(x 3) (x+ 4) =
x 3 +
x+ 4= (x+ 4) + (x 3)
(x 3) (x+ 4) =x ( + ) + (4 3)
(x 3) (x+ 4)
t ta c h :
{ + = 8
4 3 = 17 {
= 1
= 7
Vy :
8x 17
(x 3) (x+ 4) dx =
1
x 3 dx+
7
x+ 4dx = ln |x 3|+ 7 ln |x+ 4|+ c
Bi 3 : Tm nguyn hm I3 =
7x 26
x2 6x 16 dx
Gi s rng :
7x 26x2 6x 16 =
7x 26(x 8) (x+ 2) =
x 8 +
x+ 2=
( + )x+ (2 8)(x 8) (x+ 2)
Cn bng h s 2 v m ta c h :
{ + = 7
2 8 = 26 {
= 3
= 4
Vy :
7x 26
x2 6x 16 dx =
3
x 8 dx+
4
x+ 2dx = 3 ln |x 8|+ 4 ln |x+ 2|+ c
Bi 4 : Tm nguyn hm I4 =
7x2 + 75x 150
x3 25x dx
Gi s rng :
7x2 + 75x 150x3 25x =
7x2 + 75x 150x (x 5) (x+ 5) =
x+
x 5 +
x+ 5
10
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
7x2 + 75x 150
x (x 5) (x+ 5) =(x2 25)+ (x2 + 5x)+ (x2 5x)
x (x 5) (x+ 5)=
( + + )x2 + (5 5)x 25x (x 5) (x+ 5)
Cn bng h s 2 v m ta c h :
+ + = 7
5 5 = 7525 = 150
= 6
= 8
= 7Vy :
7x2 + 75x 150
x3 25x dx =
6
xdx+
8
x 5 dx
7
x+ 5dx
= 6 ln |x|+ 8 ln |x 5| 7 ln |x+ 5|+ cBi 5 : Tm nguyn hm I5 =
2x2 14x 49x3 7x2 dx
Gi s rng :
2x2 14x 49x3 7x2 =
2x2 14x 49x2 (x 7) =
x+
x2+
x 7=
(x2 7x)+ (x 7) + x2 = ( + )x2 + (7 + )x 7Cn bng h s 2 v ta c h
:
+ = 27 + = 147 = 49
= 3
= 7
= 5Vy :
2x2 14x 49x3 7x2 dx =
3
xdx+
7
x2dx
5
x 7 dx = 3 ln |x| 7
x 5 ln |x 7|+ c
Bi 6 : Tm nguyn hm I6 =
8x 36(x 5)2
dx
Gi s rng :
8x 36(x 5)2
=
x 5 +
(x 5)2= (x 5) +
(x 5)2=x 5 +
(x 5)2
Cn bng h s 2 v ta c h :
{ = 8
5 + = 36 {
= 8
= 4
Vy :
8x 36(x 5)2
dx =
8
x 5 dx+
4
(x 5)2dx = 8 ln |x 5| 4
x 5 + c
Bi 7 : Tm nguyn hm I7 =
6x2 + 20x+ 9
x3 + 2x2 + xdx
Gi s rng :
6x2 + 20x+ 9
x3 + 2x2 + x=
6x2 + 20x+ 9
x(x+ 1)2=
x+ 1+
(x+ 1)2+
x
=x (x+ 1) + x+ (x+ 1)2
x(x+ 1)2=
( + )x2 + ( + + 2)x+
x(x+ 1)2
Cn bng h s 2 v ta c h :
+ = 6
+ + 2 = 20
= 9
= 3 = 5
= 9
Vy :
6x2 + 20x+ 9
x3 + 2x2 + xdx =
3
x+ 1dx+
5
(x+ 1)2dx+
9
xdx
= 3 ln |x+ 1| 5x+ 1
+ 9 ln |x|+ 1
11
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Bi 8 : Tm nguyn hm I8 =
9x3 20x2 + 30x 97(x 2) (x 3) (x2 + 5) dx
Gi s rng :
9x3 20x2 + 30x 97(x 2) (x 3) (x2 + 5) =
x 2 +
x 3 +x+
x2 + 5
= limx2
9x3 20x2 + 30x 97(x 3) (x2 + 5) = 5 , = limx3
9x3 20x2 + 30x 97(x 2) (x2 + 5) = 4
Do :
9x3 20x2 + 30x 97(x 2) (x 3) (x2 + 5) =
5
x 2 +4
x 3 +x+
x2 + 5
Cho x = 0 ta c : 9730
= 52 4
3+
1
5
x = 1 ta c : 132
= 5 2 + 16
(+ )
Cn bng cc h s 2 v ta c h :
{ 15 =
3
5+ = 3
{
= 0
= 3
Do ta c :
9x3 20x2 + 30x 97(x 2) (x 3) (x2 + 5) =
5
x 2 +4
x 3 +3
x2 + 5
Vy :
9x3 20x2 + 30x 97(x 2) (x 3) (x2 + 5) dx =
5
x 2 dx+
4
x 3 dx+
3
x2 + 5dx
= 5 ln |x 2|+ 4 ln |x 3|+ 35
arctan
(x5
)+ c
Bi 9 : Tm nguyn hm I9 =
4x3 21x2 + 48x+ 17
(x 3)2 (x2 + 7)dx
Ta gi s rng :
4x3 21x2 + 48x+ 17(x 3)2 (x2 + 7)
=
x 3 +
(x 3)2+x+
x2 + 7
= limx3
[4x3 21x2 + 48x+ 17
x2 + 7
]= 0 , = lim
x3
[4x3 21x2 + 48x+ 17
x2 + 7
]= 5
Do :
4x3 21x2 + 48x+ 17(x 3)2 (x2 + 7)
=0
x 3 +5
(x 3)2+x+
x2 + 7=
5
(x 3)2+x+
x2 + 7
Cho x = 0 ta c :17
63=
5
9+
1
7 = 2
x = 1 ta c :3
2=
5
4+
1
8(+ ) + = 2 = 2 = 4
Suy ra :
4x3 21x2 + 48x+ 17(x 3)2 (x2 + 7)
=5
(x 3)2+
4x 2x2 + 7
Do :
4x3 21x2 + 48x+ 17
(x 3)2 (x2 + 7)dx =
5
(x 3)2dx+
4x 2x2 + 7
dx
=
5
(x 3)2dx+
4x
x2 + 7dx 2
dx
x2 + 7
= 5x 3 + 2 ln
(x2 + 7
) 27
arctan
(x7
)+ c
Ta cn ch nhng vn nh sau :
Gi s ta c hm phn thc :
P (x)
Q (x)trong bc P (x) nh hn bc ca Q (x) v ta gi s rng :
Q (x) = K (x) .R (x) = K (x) .(x a)n
12
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Do :
P (x)
Q (x)=
P (x)
K (x) .(x a)n =T (x)
K (x)+
(x a)n +
(x a)n1+ ...+
x aTrong bc ca T (x) nh hn bc ca K (x). tm cch h s , , ..., ta
lm nh sau :
= limxa
P (x)
K (x), = lim
xa
[P (x)
K (x)
], = lim
xa
[P (x)
K (x)
]nMt vn na khi phn tch mt phn thc l cc phn thc m khi mu c nghim
phc .
Cho phn thc c dng :
P (x)
Q (x).Trong : Q (x) = K (x) .R (x) = K (x) .
(ax2 + bx+ c
)Vi : = b2 4ac < 0 ta c cc nghim nh sau : xk = i, k = 0, 1,
i2 = 1Do ta c :
P (x)
Q (x)=
P (x)
K (x) . (ax2 + bx+ c)=T (x)
K (x)+
x+
ax2 + bx+ c tm cc h s : , ta lm nh sau :
= limx+i
[P (x)
K (x)
]=
Im [+ i]
, = Rec [+ i] + || .Im [+ i]
Cc th d mu nh sau :
Tm cc h s , , , :x2 + x+ 1
(x2 + 4x+ 6) (x2 + 2x+ 6)=
x+
x2 + 4x+ 6+
x+
x2 + 2x+ 6Ta c nghim ca phng trnh : x2 + 4x+ 6 = 0 x = 22 iNghim
ca phng trnh : x2 + 2x+ 6 = 0 x = 15 i
limx2+2i
[x2 + x+ 1
x2 + 2x+ 6
]=
2
3 5
2
12i
Do ta c :
=Im
[2
3 5
2
12i
]
2= 5
12,
= Rec[
2
3 5
2
12i
]+ 2.
Im
[2
3 5
2
12i
]
2=
2
3 10
12= 2
12
Tng t ta cng c : limx1+5i
[x2 + x+ 1
x2 + 4x+ 6
]= 1
12+
5
5
12i
=Im
[ 1
12+
5
5
12i
]
5=
5
12,
= Rec[ 1
12+
5
5
12i
]+
Im
[ 1
12+
5
5
12i
]
5= 1
12+
5
12=
4
12
Vy ta c c :
x2 + x+ 1
(x2 + 4x+ 6) (x2 + 2x+ 6)=
5x 212 (x2 + 4x+ 6)
+5x+ 4
12 (x2 + 2x+ 6)
Tm cc h s , , , :x2 + 1
(x2 + 6) (x+ 2)2=x+
x2 + 6+
x+ 2+
(x+ 2)2
Nghim ca phng trnh : x2 + 6 = 0 x = 6 i
13
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Ta c : limx6i
[x2 + 1
(x+ 2)2
]=
1
10+
6
5i
Do m ta c :
=Im
[1
10+
6
5i
]
6=
6
56
=1
5
= Rec[
1
10+
6
5i
]+ 0.
Im
[1
10+
6
5i
]
6=
1
10
= limx2
[x2 + 1
x2 + 6
]=
4 + 1
4 + 6=
5
10=
1
2
= limx2
[x2 + 1
x2 + 6
]= lim
x2
[10x
(x2 + 6)2
]= 20
100= 1
5
Vy ta c c :
x2 + 1
(x2 + 6) (x+ 2)2=
2x+ 1
10 (x2 + 6) 1
5 (x+ 2)+
1
2(x+ 2)2
Bi 10 : Tm nguyn hm I10 =
x+ 1
(x2 x+ 1) (x2 + 1) dx
Gi s rng :
x+ 1
(x2 x+ 1) (x2 + 1) =x+
x2 x+ 1 +x+
x2 + 1
Nghim ca phng trnh : x2 x+ 1 = 0 x = 12
3
2i
Ta c : limx 1
2+32i
[x+ 1
x2 + 1
]=
3
2
3
2i
Do m ta c :
=Im
[3
2
3
2i
]
3
2
= 1
= Rec[
3
2
3
2i
]+ |1| 1
2=
3
2+
1
2= 2
Ta c nghim ca phng trnh : x2 + 1 = 0 x = iLi c : lim
xi
[x+ 1
x2 x+ 1]
= 1 + iDo suy ra :
= Im [1 + i] = 1 , = Rec [1 + i] = 1
Do ta c :
x+ 1
(x2 x+ 1) (x2 + 1) =x+ 2
x2 x+ 1 +x 1x2 + 1
Vy :
x+ 1
(x2 x+ 1) (x2 + 1) dx =
14
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
= 12
2x 1
x2 x+ 1 dx+3
2
dx
x2 x+ 1 +1
2
2x
x2 + 1dx
dx
x2 + 1
= 12
d(x2 x+ 1)x2 x+ 1 +
3
2
dx(
x 12
)2+
3
4
+1
2
d(x2 + 1
)x2 + 1
dx
x2 + 1
= 12
ln(x2 x+ 1)+3 arctan(2x 1
3
)+
1
2ln(x2 + 1
) arctanx+ c=
1
2ln
(x2 + 1
x2 x+ 1
)+
3 arctan
(2x 1
3
) arctanx+ cBi tp bn c t luyn :
Bi 1 : Tm nguyn hm I1 =
x2
(x 1) (x+ 1) (x+ 2) dx
Bi 2 : Tm nguyn hm I2 =
x2 1(
x2 3x+ 2) (x2 1)2 dxBi 3 : Tm nguyn hm I3 =
1
(x2 + x+ 2) (x2 + 1)2dx
Bi 4 : Tm nguyn hm I4 =
x2 1
(x3 + 2) (x 1)3dx
PHNG PHP LNG GIC HA TRONG TCH PHN.
Dng tch phn i bin s iu kin bin s
f(x,a2 x2) dx x = a sin t t [pi
2;pi
2
]f(x,x2 a2) dx x = a
cos tt [0;pi
2
)[pi;
3pi
2
)f(x,x2 + a2
)dx x = a tan t t
[0;pi
2
)f
(x,
a+ x
a x)dx x = a cos 2t t
(0;pi
2
)f(x,
(x a) (b x))dx x = a+ (b a) sin2t t
[0;pi
2
]Bng 1: Bng tm tt cc cng thc bin i lng gic bng cch i bin s.
1. Dng 1 : f(x,a2 x2) dxBi 1 : Tm nguyn hm A1 =
xdx1 x2t : x = sin t dx = cos tdtVy :
xdx1 x2 =
sin t. cos tdt
1 sin2t=
sin t. cos tdt
|cos t|
=
sin t. cos tdt
cos t=
sin tdt = cos t+ c, 0 t < pi
2,
sin t. cos tdt
cos t=
sin tdt = cos t+ c, pi t < 3pi
2,
15
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Cch 2 :
iu kin : 1 < x < 1,Bin i :
xdx1 x2 =
xdx
(1 x) (1 + x)=
xdx
1 + x.
1 xTa nhn thy : 1 x+ 1 + x = 2. t : 1 x =
2 sin t,
1 + x =
2 cos t
Do ta c : 1 x = 2sin2t dx = 4 sin t. cos tdtVy :
xdx
1 + x.
1 x = 4 (
1 2sin2t) sin t. cos tdt2 sin t. cos t
= 2
cos 2tdt = sin 2t+ cCch 3 :
Ta c :
xdx1 x2 =
1
2
d(1 x2)1 x2 =
1
2
(1 x2) 12d (1 x2) = 1 x2 + cBi 2 : Tm nguyn hm A2 =
dx
2 + x+
2 xTa ch : 2 + x+ 2 x = 4,Do ta t :
2 + x = 2 sin t,
2 x = 2 cos t, 0 t pi
2;
2 + x = 4sin2t dx = 8 sin t cos tdtVy :
dx
2 + x+
2 x = 4
sin t cos tdt
sin t+ cos t= 2
(sin t+ cos t)2 1
sin t+ cos tdt
= 2
(sin t+ cos t)dt 2
dt
sin t+ cos t= 2
(sin t+ cos t)dt
2
dt
sin(t+
pi
4
)= 2
(sin t+ cos t)dt
2
dt
2 sin(t
2+pi
8
). cos
(t
2+pi
8
)= 2 (sin t cos t)
2 lntan( t
2+pi
8
)+ cBi 3 : Tnh tch phn A3 =
32
0
(1 x2)1 x2 dxt : x = sin t dx = cos tdt. i cn : Vi x = 0 t = 0,
x =
3
2 t = pi
3
Vy : A3 =
pi3
0
(1 sin2t)1 sin2t cos tdt =
pi3
0
(1 sin2t)cos2tdt =
pi3
0
cos4tdt
=1
4
pi3
0
(1 + cos 2t)2dt =1
4
pi3
0
(1 + 2 cos 2t+ cos22t
)dt
=1
4
pi3
0
(1 + 2 cos 2t+
1 + cos 4t
2
)dt =
1
8
pi3
0
(3 + 4 cos 2t+ cos 4t) dt
=1
8
(3t+ 2 sin 2t+
1
4sin 4t
) pi3
0
=pi
8+
7
3
64
Cch 2 :
16
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Vi iu kin : 0 x
3
2ta c :
1 x2 = 1 x.1 + xCh rng : 1 x+ 1 + x = 2Do ta t :
1 x =
2 sin t,
1 + x =
2 cos t, dx = 4 sin t. cos tdti cn : x = 0 t = pi
4, x =
3
2 t = pi
12
Vy : A3 = 32pi12
pi4
sin2t.cos2t. sin t. cos t. sin t. cos tdt = 32pi12
pi4
sin4t.cos4tdt
= 2pi12
pi4
sin42tdt = 2pi12
pi4
(1 cos 4t
2
)2dt = 1
2
pi12
pi4
(1 2 cos 4t+ 1 + cos 8t
2
)dt
= 12
(3
2t 1
2sin 4t+
1
16sin 8t
) pi12
pi4
=pi
8+
7
3
64
Bi 4 : Tnh tch phn A4 =
10
dx
(2 x2)2 x2
t : x =
2 sin t dx =
2 cos tdt. i cn : x = 0 t = 0, x = 1 t = pi4,
Vy : A4 =
pi4
0
2 cos tdt(
2 2sin2t)2 2sin2t =pi4
0
2 cos tdt(
2 2sin2t)2 cos t = 12pi4
0
dt
cos2t
=1
2tan t
pi4
0
=1
2
Cch 2 :
Vi iu kin : 0 x 1 ta c :
2 x2 =(
2 x) (2 + x) =2 x.2 + xt :
(2 x) =22 sin t,(2 + x) =22 cos t,Do :
2 x = 2
2 sin t, dx = 4
2 sin t. cos tdt
i cn : x = 0 t = pi4, x = 1 t = pi
8,
Vy : A4 =
pi8
pi4
42 sin t. cos tdt8sin2t.cos2t.2
2 sin t. cos t
= 14
pi8
pi4
dt
sin2t.cos2t= 1
4
pi8
pi4
(1 +
1
tan2t
)d (tan t)
= 14
(tan t 1
tan t
) pi8
pi4
=1
2 0 = 1
2
Bi 5 : Tnh tch phn A5 =
0 1
2
dx
1 +x (x+ 1)
Ta c : x (x+ 1) = (x2 + x) = (x2 + 2.12.x+
1
4 1
4
)=
1
4(x+
1
2
)2
17
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
t : x+1
2=
1
2sin t dx = 1
2cos tdt. i cn : x = 1
2 t = 0, x = 0 t = pi
2,
Vy :A5 =
0 1
2
dx
1 +
1
4(x+
1
2
)2 =pi2
0
1
2cos tdt
1 +
1
4 1
4sin2t
=
pi2
0
cos tdt
2 + cos t=pi
22
pi2
0
dt
2 + cos t
=pi
2 2
pi2
0
dt
1 + 2cos2t
2
=pi
2 4
pi2
0
d(
tant
2
)tan2
t
2+ 3
=pi
2 4
3arctan
tan t23
pi2
0
=pi
2+
43
arctan
(03
) 4
3arctan
(13
)=pi
2 2pi
3
3=
9 4318
pi
Cch 2 :
Ta ch : x+ x+ 1 = 1 . t : x = sin t,1 + x = cos t x = sin2t dx =
2 sin t. cos tdt. i cn : x = 1
2 t = pi
4, x = 0 t = 0
Vy :
0 1
2
dx
1 +x (x+ 1)
=
0pi4
2 sin t. cos tdt1 + sin t. cos t
= 0
pi4
(2 + 2 sin t. cos t 2) dt1 + sin t. cos t
=pi
2 2
pi4
0
dt
1 + sin t. cos t=pi
2 2
pi4
0
d (tan t)
tan2t+ tan t+ 1
=pi
2 2
pi4
0
d (tan t)(tan t+
1
2
)2+
3
4
=pi
2 2J
t : u = tan t, Vi : t = 0 u = 0, t = pi4 u = 1
J =pi4
0
d (tan t)(tan t+
1
2
)2+
3
4
=
10
dt(t+
1
2
)2+
3
4
t : t+1
2=
3
2tan v dt =
3
2
(1 + tan2v
)dv
i cn : t = 0 v = pi6, t = 1 v = pi
3,
Do :
10
dt(t+
1
2
)2+
3
4
=
pi3
pi6
3
2
(1 + tan2v
)dv
3
4tan2v +
3
4
=23
pi3
pi6
dv =23v
pi3
pi6
=pi
3
3
Vy ta c c : I =pi
2 2J = pi
2 2pi
3
3=
9 4318
pi
2. Dng 2 : f(x,x2 a2) dxBi 1 : Tm nguyn hm B1 =
xx2 1 dx
18
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
t : x =1
cos t, t
[0;pi
2
)[pi;
3pi
2
), dx = sin t
cos2tdt
Vy : B1 =
1cos2t
1cos t
.sin t
cos2tdt =
sin2tcos2t
cos t.
sin t
cos2tdt =
sin t. sin tcos t
cos3tdt
=
sin2t
cos4tdt =
1 cos2t
cos4tdt =
1 cos2t
cos2td (tan t)
=
(1 + tan2t 1) d (tan t) = tan3t
3+ c
Bi 2 : Tnh tch phn B2 =
42
x2x2 4 dx
t : x =2
cos t dx = 2 sin t
cos2tdt, i cn : Vi gi tr x = 2 t = 0, x = 4 t = pi
3
Vy : B2 =
pi3
0
4
4
cos2t 4
cos2t.2 sin t
cos2tdt = 16
pi3
0
4
sin2t
cos2tcos2t
.2 sin t
cos2tdt = 16
pi3
0
sin2t
cos5tdt
= 16
pi3
0
sin2t
cos6td (sin t) = 16
pi3
0
sin2t(1 sin2t)3 d (sin t)Li t : u = sin t, t = 0 u = 0, t =
pi
3 u =
3
2
Do :
sin2t(1 sin2t)3 = u
2
(1 u2)3= u
2
(u 1)3(1 + u)3
Phn tch : 16u2
(1 u)3(1 + u)3=
u 1 +
(u 1)2+
(u 1)3+
1 + u+
(1 + u)2+
(1 + u)3
= limx1
[ 16u
2
(1 + u)3
](2)= 1, = lim
x1
[ 16u
2
(1 + u)3
],= 1, = lim
x1
[ 16u
2
(1 + u)3
]= 2
= limx1
[ 16u
2
(u 1)3](2)
= 1, = limx1
[ 16u
2
(u 1)3],
= 1,
= limx1
[ 16u
2
(u 1)3]
= 2
Suy ra : 16u2
(1 u)3(1 + u)3=
1
u 1 1
(u 1)2 2
(u 1)3 1
(1 + u) 1
(1 + u)2+
2
(1 + u)3
Vy : 16
pi3
0
sin2t(1 sin2t)3d (sin t) =
=
32
0
[1
u 1 1
(u 1)2 2
(u 1)3 1
(1 + u) 1
(1 + u)2+
2
(1 + u)3
]du
19
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
= 16
32
0
u2
(u 1)3(1 + u)3du =
[2(u3 + u
)(u2 1)2
+ lnu 1u+ 1
] 32
0
= 28
3+ln
(232 +
3
)Cch 2 :
t : t =x2 4 t2 = x2 4 tdt = xdx dx
t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 4 (t x) (t+ x) = 4Ta li t : u = t+
x t x = 4
ui cn : x = 2 t = 2, x = 4 t = 4 + 2
3
Vy :
x2x2 4dx = 1
16
(u+
4
u
)2(u 4
u
)2duu
=
(16
u5+u3
16 2
4
)du
= 4u4
+u4
64 2 ln |u|+ c
Thay cn vo ta c : B2 =
[ 4u4
+u4
64 2 ln |u|
] 4+23
2
= 28
3 2 ln (2 +3)Bi 3 : Tnh tch phn B3 =
223
dx
(x2 1)x2 1
t : x =1
cos t dx = sin tdt
cos2t.i cn : Vi gi tr x =
23 t = pi
6, x = 2 t = pi
3
Vy : B3 =
pi3
pi6
sin tdt
cos2t(
1
cos2t 1)
1
cos2t 1
=
pi3
pi6
cos tdt
sin2t=
pi3
pi6
d (sin t)
sin2t
= 1sin t
pi3
pi6
=6 23
3
Cch 2 :
t : t =x2 1 t2 = x2 1 tdt = xdx, dt
x=dx
t
Do ta suy ra :
dx
t=dx+ dt
x+ t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 1 (t+ x) (t x) = 1t : u = t+ x t x
= 1
u,Vi gi tr x =
23 u =
3, x = 2 u = 2 +
3
Vy : B3 = 4
2+3
3
du(u 1
u
)2u
= 4
2+3
3
udu
(u2 1)2=2
2+3
3
d(u2 1)
(u2 1)2
=2
1 u2
2+3
3
=6 23
3
Bi 4 : Tnh tch phn B4 =
0 1
2
dx
(x+ 1)
3 2x x2
20
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Ta c : B4 =
0 1
2
dx
(x+ 1)
4 (x+ 1)2
= 0
12
d(
1
x+ 1
)2
1
(x+ 1)2 1
4
=
21
ds
2
s2 1
4
Ch : Nu ta t : t =1
2 cos sth khi i cn th s ra hm s ngc arc... , do m ta
chn cch gii nh sau :
t : t =
s2 1
4 tdt = sds ds
t=d (s+ t)
s+ t
Do cch t ta c : (t s) (t+ s) = 1, t : u = t+ s t s = 1u
i cn : s = 1 u = 1 +
3
2, s = 2 u = 2 +
15
2
Do , ta c :
ds
2s2 14
=
du
2u=
1
2ln |u|+ c
Thay cn vo ta c : B4 =1
2ln |u|
2+
152
1+32
=1
2ln
(4 +
15
2 +
3
)
Bi 5 :Tnh tch phn B5 =
0 1
2
dx
(x+ 1)
3 + 2x x2
Bin i :
0 1
2
dx
(x+ 1)
3 + 2x x2 =0
12
dx
(x+ 1)
(x+ 1) (3 x)=
0 1
2
dx
(x+ 1)2
3 xx+ 1
t : t =
3 xx+ 1
tdt2 =dx
(x+ 1)2
i cn: x = 12 t =
7, x = 0 t =
3
Vy :
0 1
2
dx
(x+ 1)2
3xx+1
=
7
3
tdt
2t=
73
2
Cch 2 :
Ta c :
0 1
2
dx
(x+ 1)
3 + 2x x2 =0
12
dx
(x+ 1)
4 (x 1)2
Gi s rng : i2 = 1t : t =
4 (x 1)2 t2 i2(x 1)2 = 4 (t ix+ i) (t+ ix i) = 4Do : tdt = i (x
1) d (ix) dix
t=
dt
i (x 1) =d (ix+ t i)t+ ix iLi t : u = t+ ix i ,do m ta c : t ix+
i = 4
u
Vy ta c : t =1
2
(u+
4
u
), x =
1
2i
(u 4
u
)+ 1
21
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Suy ra :
d (ix)
i (x+ 1)
4 (x 1)2
=
d (ix)
i (x+ 1) t=
du(
1
2
(u 4
u
)+ 2i
)u
= 2
du
u2 4 + 4ui
= 2
du
u2 + 4ui+ 4i2= 2
d (u+ 2i)
(u+ 2i)2= 2
u+ 2i+ c
Vi gi tr : x = 12 u =
7
2 3
2i;x = 0 u =
3 i
Thay cn vo ta c c : B5 = 2u+ 2i
3i
72 3
2i
=
3
2+
1
2i+
7
2 1
2i =
73
2
3. Dng 3 : f(x,x2 + a2
)dx
Bi 1 : Tnh tch phn C1 =
10
x2 + 1 dx
t : x = tan t dx = (1 + tan2t) dti cn : x = 0 t = 0, x = 1 t =
pi
4
Do :
10
x2 + 1dx =
pi4
0
tan2t+ 1
(1 + tan2t
)dt =
pi4
0
dt
cos3t=
pi4
0
d (sin t)
cos4t
=
pi4
0
d (sin t)(1 sin2t)2 = 14
pi4
0
(1 sin t+ 1 + sin t(1 sin t) (1 + sin t)
)2d (sin t)
=1
4
pi4
0
(1
1 + sin t+
1
1 sin t)2d (sin t)
=1
4
pi4
0
(1
1 + sin t
)2d (sin t)+
1
4
pi4
0
(1
1 sin t)2d (sin t)+
1
2
pi4
0
(1
(1 + sin t) (1 sin t)
)d (sin t)
=
[ 1
4 (1 + sin t)+
1
4 (1 sin t)
] pi4
0
+1
4
pi4
0
(1 + sin t+ 1 sin t(1 + sin t) (1 sin t)
)d (sin t)
=
2
2+
1
4ln
1 + sin t1 sin tpi4
0
=
2
2+
1
2ln(1 +
2)Cch 2 :
t : t =x2 + 1 t2 = x2 + 1 tdt = xdxDo :
dx
t=
(x+ t) dx
(x+ t) t=xdx+ tdx
(x+ t) t=tdt+ tdx
(x+ t) t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 + 1 (t+ x) (t x) = 1Li t : u = t+
x t x = 1
ui cn : x = 0 t = 1, x = 1 t = 1 +2
22
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Do :
x2 + 1dx =
(x2 + 1
)dx
x2 + 1=
1
4
(u+ 1u
)2du
u=
(u+
2
u+
1
u3
)du
=u2
8+
1
2ln |u| 1
8u2+ c
Thay cn vo ta c c :
(u2
8+
1
2ln |u| 1
8u2
) 1+2
1
=
2
2+
1
2ln(1 +
2)Bi 2 : Tnh tch phn C2 =
3
1
1
x2ln
1 + x2dx
t : x = tan t dx = (1 + tan2t) dti cn : x = 1 t = pi
4, x =
3 x = pi
3
Vy :
3
1
1
x2ln
1 + x2dx =
pi3
pi4
1
tan2tln
1 + tan2t(1 + tan2t
)dt
= pi3
pi4
(cot2x+ 1
)ln (cos t) dt =
pi3
pi4
ln (cos t) d (cot t)
= cot t ln (cos t)
pi3
pi4
+
pi3
pi4
cot t.sin t
cos tdt =
3
3ln(
1
2
) ln
(2
2
)+pi
12
Cch 2 :
t : t =x2 + 1 t2 = x2 + 1, tdt = xdxDo m ta c :
dx
t=
(x+ t) dx
(x+ t) t=xdx+ tdx
(x+ t) t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 + 1 (t+ x) (t x) = 1Li t : u = t+
x t x = 1
ui cn : x = 1 t = 1 +
2, x =
3 t = 2 +
3
Vy :
1
x2ln
1 + x2dx = 2
(u+ 1u
)(u 1
u
)2u
ln(
1
2
(u+
1
u
))du
= 2
ln(
1
2
(u+
1
u
))d
1u 1
u
= 2ln(
1
2
(u+
1
u
))u 1
u
+ 2
1 1u2
u2 1u2
du+ c
= 2ln(
1
2
(u+
1
u
))u 1
u
+ 2
du
u2 + 1+ c = 2
ln(
1
2
(u+
1
u
))u 1
u
+ 2 arctanu+ c
Thay cn vo ta c :
23
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
C2 =
2ln(
1
2
(u+
1
u
))u 1
u
+ 2 arctanu
2+3
1+2
= ln 23
+ ln
2 + 2 arctan(2 +
3) 2 arctan (1 +2)
=
3
3ln(
1
2
) ln
(2
2
)+
pi
12
Bi 3 : Tnh tch phn C3 =
43
1
x2 + 1
xx4 + 1
dx
Bin i :
43
1
x2 + 1
xx4 + 1
dx =1
2
43
1
(x2 + 1
)d(x2)
x2x4 + 1
t : x2 = tan t d (x2) = (1 + tan2t) dti cn : x = 1 t = pi
4, x =
4
3 t = pi3
Vy : C3 =1
2
pi3
pi4
(tanx+ 1) d (tanx)
tanx
tan2x+ 1=
1
2
pi3
pi4
(tanx+ 1) cos2x(1 + tan2x
)dx
sinx
=1
2
pi3
pi4
(tanx+ 1) dx
sinx=
1
2
pi3
pi4
(1
cosx+
1
sinx
)dx = 1
2
pi3
pi4
d (sinx)
sin2x 1+1
2
pi3
pi4
d (cosx)
cos2x 1
=1
4
[ ln
sinx 1sinx+ 1+ ln cosx 1cosx+ 1
] pi3
pi4
=1
4
[2 ln (23)+ ln (3 22) ln 3 ln (3 22)]=
1
4
[2 ln (23) ln 3]Bi 4 : Tnh tch phn C4 =
10
x3x2 + 1 dx
t : x = tan t dx = (1 + tan2t) dti cn : x = 0 t = 0, x = 1 t =
pi
4
Vy : C4 =
pi4
0
tan3t
tan2t+ 1(1 + tan2t
)dt =
pi4
0
sin3t
cos6tdt =
pi4
0
(cos2t 1) d (cos t)
cos6t
=
pi4
0
(1
cos4t 1
cos6t
)d (cos t) =
[ 1
3cos3t+
1
5cos5t
] pi4
0
= 2
2
3+
4
2
5+
2
15=
2 + 2
2
15Cch 2 : t =
x2 + 1 t2 = x2 + 1, tdt = xdx
24
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Do :
dx
t=
(x+ t) dx
(x+ t) t=xdx+ tdx
(x+ t) t=tdt+ tdx
(x+ t) t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 + 1 (t+ x) (t x) = 1Li t : u = t+
x t x = 1
uVi gi tr : x = 0 u = 1, x = 1 u = 1 +2Vy :
x3x2 + 1dx =
1
32
(u 1
u
)3(u+
1
u
)2duu
=1
32
(u 1
u
)2(u+
1
u
)2 (1 1
u2
)du
=1
32
(u 1
u
)2(u+
1
u
)2d(u+
1
u
)=
1
32
[(u+
1
u
)2 4](
u+1
u
)2d(u+
1
u
)=
[1
160
(u+
1
u
)5 1
24
(u+
1
u
)3]+ c
Thay cn vo ta c :
C4 =
[1
160
(u+
1
u
)5 1
24
(u+
1
u
)3] 1+2
1
=
(2
2)5
160(2
2)3
24+
2
15=
2 + 2
2
15
Bi 5 : Tnh tch phn C5 =
pi4
0
dx
cosx
sin 2x+ 2
Phn tch : sin 2x+ 2 = 2 sin x cosx+ 2 = cos2x(
2 tanx+2
cos2x
)= 2cos2x
(tan2x+ tanx+ 1
)= 2cos2x
[(tanx+
1
2
)2+
3
4
]
Do :
pi4
0
dx
cosx
sin 2x+ 2=
pi4
0
dx
2cos2x
(tanx+ 12
)2+ 34
t : tanx+1
2=
3
2tan t (1 + tan2x) dx = 3
2
(1 + tan2t
)dt
i cn : x = 0 t = pi6, x =
pi
4 t = pi
3
Vy : C5 =
pi3
pi6
3(1 + tan2t
)dt
2
2
3
4tan2t+
3
4
=
pi3
pi6
(1 + tan2t
)dt
2
tan2t+ 1=
12
pi3
pi6
dt
cos t= 1
2
pi3
pi6
d (sin t)
sin2t 1
= 12
2ln
sin t 1sin t+ 1pi3
pi6
= 12
ln(2
3) 1
2
2ln 3
Bi 6 : Tnh tch phn C6 =
33
0
dx(x2 + 9)
3
t : x = 3 tan t dx = 3 (1 + tan2t) dt , Vi : x = 0 t = 0, x = 33
t = pi3
25
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
Vy : C6 =
pi3
0
(1 + tan2t
)dt
9(1 + tan2t
)1 + tan2t
=1
9
pi3
0
cos tdt =1
9sin t
pi3
0
=
3
18
Cch 2 :
t : t =x2 + 9 t2 = x2 + 9, tdt = xdxDo :
dx
t=
(x+ t) dx
(x+ t) t=xdx+ tdx
(x+ t) t=dx+ dt
x+ t=d (x+ t)
x+ tDo cch t m ta c : t2 = x2 + 9 (t+ x) (t x) = 9Li t : u = t+
x t x = 9
ui cn : x = 0 t = 3, x = 33 u = 6 + 33Vy :
dx
(x2 + 9)3
= 4
du(
u+ 9u)2u
= 2
d(u2 + 9
)(u2 + 9)
2= 2
u2 + 9+ c
Thay cn vo ta c : C6 = 2u2 + 9
6+33
3
=4 + 23
36+
2
18=
3
18
4. Dng 4 : f
(x,
a+ x
a x)dx
Bi 1 : Tnh tch phn D1 =
12
0
x
1 + x
1 x dx
t : x = cos 2t dx = 2 sin 2tdti cn : x = 0 t = pi
4, x =
1
2 t = pi
6
Vy : D1 = 2pi6
pi4
cos 2t. sin 2t
1 + cos 2t
1 cos 2t dt = 2pi6
pi4
cos 2t. sin 2t
2cos2t
2sin2tdt
= 4pi6
pi4
cos 2t. sin t. cos tcos t
sin tdt = 4
pi6
pi4
cos 2t.cos2tdt = 2pi6
pi4
cos 2t. (1 + cos 2t) dt
= 2pi6
pi4
(cos 2t+
1 + cos 4t
2
)dt =
[ sin 2t 1
4sin 4t t
] pi6
pi4
=53 + 8
8+
pi
12
Cch 2 :
Ta nhn thy : 1 + x+ 1 x = 2 do t : 1 + x =
2 sin t,
1 x =
2 cos t
Suy ra : 1 + x = 2sin2t dx = 4 sin t. cos tdti cn : x = 0 t =
pi
4, x =
1
2 t = pi
3
Vy : D1 = 4
pi3
pi4
(2sin2t 1) sin t
cos tsin t. cos tdt =4
pi3
pi4
(2sin2t 1) sin2tdt
26
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
= 2
pi3
pi4
cos 2t (cos 2t 1) dt = 2pi3
pi4
(1 + cos 4t
2 cos 2t
)dt
=[
1
4sin 4t sin 2t+ t
] pi3
pi4
=53 + 8
8+
pi
12
Bi 2 : Tnh tch phn D2 =
30
xdx3 + x+
3 x
Bin i :
30
xdx3 + x+
3 x =
30
x3 + x
1 +
3 x3 + x
dx
t : x = 3 cos 2t dx = 6 sin 2tdti cn : x = 0 t = pi
4, x = 3 t = 0
Do : D2 = 6
pi4
0
3 cos 2t3 + 3 cos 2t
1 +
3 3 cos 2t3 + 3 cos 2t
sin 2tdt = 6
pi4
0
3 cos 2t6cos2t
1 +
6sin2t
6cos2t
sin 2tdt
= 6
6
pi4
0
cos 2t
sin t+ cos tsin t cos tdt = 6
6
pi4
0
(cos t sin t) sin t cos tdt
= 6
6
pi4
0
sin tcos2tdt6
6
pi4
0
sin2t cos tdt = 6
6
pi4
0
cos2td (cos t)6
6
pi4
0
sin2td (sin t)
= 6
6
[cos3t
3+
sin3t
3
] pi4
0
= 2
3 + 2
6
Cch 2 :
Ta nhn thy : 3 + x+ 3 x = 6 , do t : 3 + x =
6 sin t,
3 x =
6 cos t
Rt x ta c c : x = 6sin2t 3 = 3 (2sin2t 1) , dx = 12 sin t cos
tdti cn : x = 0 t = pi
4, x = 3 t = pi
2
Vy : D2 =
pi2
pi4
36(2sin2t 1) sin t cos tdt
6 (sin t+ cos t)=
pi2
pi4
36 (sin t cos t) (cos t+ sin t) sin t cos tdt6 (sin t+ cos
t)
=
pi2
pi4
6
6 (sin t cos t) sin t cos tdt = 6
6
pi2
pi4
sin2t cos tdt 6
6
pi2
pi4
sin tcos2tdt
= 6
6
pi2
pi4
sin2td (sin t) + 6
6
pi2
pi4
cos2td (cos t)
27
-
Tuyn tp cc chuyn v k thut tnh nguyn hm,tch phn. Nguyn Duy ng
=6
6
3
[sin3t+ cos3t
] pi2
pi4
= 2
6 2
3
5. Dng 1 : f(x,
(x a) (b x))dx
Bi 1 : Tnh tch phn E1 =
21
(x 1) (2 x) dx
t : x = 1 + sin2t dx = 2 sin t. cos tdt . i cn : x = 1 t = 0, x
= 2 t = pi2
Do : E1 = 2
pi2
0
sin t. cos t
(1 + sin2t 1) (2 1 sin2t)dt
= 2
pi2
0
sin t. cos t
sin2t
(1 sin2t)dt = 2
pi2
0
sin2t.cos2tdt
=1
2
pi2
0
sin22tdt =1
4
pi2
0
(1 cos 4t)dt =[
1
4t 1
8sin 4t
] pi2
0
=pi
8
Cch 2 :
Ta ch : x 1 + 2 x = 1 , do ta t : x 1 = sin t,2 x = cos tDo : x
1 = sin2t dx = 2 sin t cos tdt .i cn : x = 1 t = 0, x = 2 t =
pi
2
Vy : E1 = 2
pi2
0
sin2t.cos2t =1
2
pi2
0
sin22tdt =1
4
pi2
0
(1 cos 4t)dt =[
1
4t 1
8sin 4t
] pi2
0
=pi
8
28
