Bài tập toán cao cấp tập 3 part 2 pdf
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10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 33
Phu
.
o
.
ng ph´ap II. Thay x = 1 v`ao (10.7) ta c´o 1 = A · 4 ⇒ A =
1
4
.
Tiˆe
´
p theo, thay x = −1 v`ao (10.7) ta thu d
u
.
o
.
.
c: −1=−B
2
· 2hay
l`a B
2
=
1
2
.D
ˆe
’
t`ım B
1
ta thˆe
´
gi´a tri
.
x = 0 v`ao (10.7) v`a thu du
.
o
.
.
c
0=A − B
1
− B
2
hay l`a B
1
= A − B
2
= −
1
4
.
Do d
´o
I =
1
4
dx
x − 1
−
1
4
dx
x +1
+
1
2
dx
(x +1)
2
= −
1
2(x +1)
+
1
4
ln
x − 1
x +1
+ C.
V´ı du
.
2. T´ınh I =
3x +1
x(1 + x
2
)
2
dx.
Gia
’
i. Khai triˆe
’
n h`am du
.
´o
.
idˆa
´
u t´ıch phˆan th`anh tˆo
’
ng c´ac phˆan
th ´u
.
cco
.
ba
’
n
3x +1
x(1 + x
2
)
2
=
A
x
+
Bx + C
1+x
2
+
Dx + F
(1 + x
2
)
2
T`u
.
d
´o
3x +1≡ (A + B)x
4
+ Cx
3
+(2A + B + D)x
2
+(C + F )x + A.
Cˆan b˘a
`
ng c´ac hˆe
.
sˆo
´
cu
’
a c´ac l ˜uy th`u
.
ac`ung bˆa
.
ccu
’
a x ta thu d
u
.
o
.
.
c
A + B =0
C =0
2A + B + D =0⇒ A =1,B = −1,C =0,D = −1,F =3
C + F =3
A =1.
T`u
.
d
´o suy r˘a
`
ng
I =
dx
x
−
xdx
1+x
2
−
xdx
(1 + x
2
)
2
+3
dx
(1 + x
2
)
2
=ln|x|−
1
2
ln(1 + x
2
) −
1
2
(1 + x
2
)
−2
d(1 + x
2
)+3
dx
(1 + x
2
)
2
=ln|x|−
1
2
ln(1 + x
2
)+
1
2(1 + x
2
)
+3I
2
.
34 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
Ta t´ınh I
2
=
dx
(1 + x
2
)
2
b˘a
`
ng cˆong th ´u
.
c truy hˆo
`
ithud
u
.
o
.
.
c trong
10.1. Ta c´o
I
2
=
1
2
·
x
1+x
2
+
1
2
I
1
=
x
2(1 + x
2
)
+
1
2
dx
1+x
2
=
x
2(1 + x
2
)
+
1
2
arctgx + C.
Cuˆo
´
ic`ung ta thu d
u
.
o
.
.
c
I =ln|x|−
1
2
ln(1 + x
2
)+
3x +1
2(1 + x
2
)
+
3
2
arctgx + C.
B
`
AI T
ˆ
A
.
P
T´ınh c´ac t´ıch phˆan (1-12)
1.
xdx
(x + 1)(x + 2)(x − 3)
.
(D
S.
1
4
ln |x +1|−
2
5
ln |x +2| +
3
20
|x − 3|)
2.
2x
4
+5x
2
− 2
2x
3
− x −1
dx.
D
S.
x
2
2
+ln|x − 1| + ln(2x
2
+2x + 1) + arctg(2x + 1))
3.
2x
3
+ x
2
+5x +1
(x
2
+ 3)(x
2
− x +1)
dx.
D
S.
1
√
3
arctg
x
√
3
+ ln(x
2
− x +1)+
2
√
3
arctg
2x − 1
√
3
)
4.
x
4
+ x
2
+1
x(x − 2)(x +2)
dx.
(D
S.
x
2
2
−
1
4
ln |x|+
21
8
ln |x − 2| +
21
8
ln |x +2|)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 35
5.
dx
x(x − 1)(x
2
− x +1)
2
.
(D
S. ln
x −1
x
−
10
3
√
3
arctg
2x − 1
√
3
−
1
3
2x − 1
x
2
− x +1
)
6.
x
4
−x
2
+1
(x
2
− 1)(x
2
+ 4)(x
2
−2)
dx.
(D
S. −
1
10
ln
x − 1
x +1
+
7
20
arctg
x
2
+
1
4
√
2
ln
x −
√
2
x +
√
2
)
7.
3x
2
+5x +12
(x
2
+ 3)(x
2
+1)
dx.
(D
S. −
5
4
ln(x
2
+3)−
√
3
2
arctg
x
√
3
+
5
4
ln(x
2
+1)+
9
2
arctgx)
8.
(x
4
+1)dx
x
5
+ x
4
−x
3
− x
2
.
(D
S. ln |x|+
1
x
+
1
2
ln |x − 1|−
1
2
ln |x +1| +
1
x +1
)
9.
x
3
+ x +1
x
4
−1
dx.
(D
S.
3
4
ln |x − 1| +
1
4
ln |x +1|−
1
2
arctgx)
10.
x
4
1 − x
4
dx.
(D
S. − x +ln
x +1
x − 1
+
1
2
arctgx)
11.
3x +5
(x
2
+2x +2)
2
dx.
(D
S.
2x − 1
2(x
2
+2x +2)
+ arctg(x + 1))
36 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
12.
x
4
− 2x
2
+2
(x
2
− 2x +2)
2
dx.
(D
S. x +
3 − x
x
2
− 2x +2
+2ln(x
2
− 2x + 2) + arctg(x − 1))
13.
x
2
+2x +7
(x − 2)(x
2
+1)
3
dx.
(D
S.
3
5
ln |x
2
− 2|−
3
10
ln |x
2
+1|+
1 − x
x
2
+1
−
11
5
arctgx)
14.
x
2
(x +2)
2
(x +1)
dx.
(D
S.
4
x +2
+ln|x +1|)
15.
x
2
+1
(x − 1)
3
(x +3)
dx.
(D
S. −
1
4(x − 1)
2
−
3
8(x − 1)
+
5
32
ln
x − 1
x +3
)
16.
dx
x
5
− x
2
(DS.
1
x
+
1
6
ln
(x −1)
2
x
2
+ x +1
+
1
√
3
arctg
2x +1
√
3
)
17.
3x
2
+8
x
3
+4x
2
+4x
dx.
(D
S. 2 ln |x| +ln|x +2| +
10
x +2
)
18.
2x
5
+6x
3
+1
x
4
+3x
2
dx.
(D
S. x
2
−
1
3x
−
1
3
√
3
arctg
x
√
3
)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 37
19.
x
3
+4x
2
− 2x +1
x
4
+ x
dx.
(D
S. ln
|x|(x
2
− x +1)
(x +1)
2
+
2
√
3
arctg
2x − 1
√
3
)
20.
x
3
− 3
x
4
+10x
2
+25
dx.
(D
S.
1
2
ln(x
2
+5)+
25 − 3x
10(x
2
+5)
−
3
10
√
5
arctg
x
√
5
)
Chı
’
dˆa
˜
n. x
4
+10x
2
+25=(x
2
+5)
2
.
10.2.2 T´ıch phˆan mˆo
.
tsˆo
´
h`am vˆo ty
’
do
.
n gia
’
n
Mˆo
.
tsˆo
´
t´ıch phˆan h`am vˆo ty
’
thu
.
`o
.
ng g˘a
.
p c´o thˆe
’
t´ınh d
u
.
o
.
.
cb˘a
`
ng phu
.
o
.
ng
ph´ap h˜u
.
uty
’
h´oa h`am du
.
´o
.
idˆa
´
u t´ıch phˆan. Nˆo
.
i dung cu
’
aphu
.
o
.
ng ph´ap
n`ay l`a t`ım mˆo
.
tph´ep biˆe
´
nd
ˆo
’
idu
.
a t´ıch phˆan d
˜a cho cu
’
a h`am vˆo ty
’
vˆe
`
t´ıch phˆan h`am h˜u
.
uty
’
. Trong tiˆe
´
t n`ay ta tr`ınh b`ay nh˜u
.
ng ph´ep d
ˆo
’
i
biˆe
´
n cho ph´ep h˜u
.
uty
’
h´oa d
ˆo
´
iv´o
.
imˆo
.
tsˆo
´
l´o
.
p h`am vˆo ty
’
quan tro
.
ng
nhˆa
´
t. Ta quy u
.
´o
.
ck´yhiˆe
.
u R(x
1
,x
2
, )hayr(x
1
,x
2
, ) l`a h`am h˜u
.
u
ty
’
d
ˆo
´
iv´o
.
imˆo
˜
ibiˆe
´
n x
1
,x
2
, ,x
n
.
I. T´ıch phˆan c´ac h`am vˆo ty
’
phˆan tuyˆe
´
n t´ınh. T´ıch phˆan da
.
ng
R
x,
ax + b
cx + d
p
1
, ,
ax + b
cx + d
p
n
dx (10.8)
trong d
´o n ∈ N; p
1
, ,p
n
∈ Q; a, b, c ∈ R; ad − bc =0du
.
o
.
.
ch˜u
.
uty
’
h´oa nh`o
.
ph´ep d
ˆo
’
ibiˆe
´
n
ax + b
cx + d
= t
m
o
.
’
d
ˆay m l`a mˆa
˜
usˆo
´
chung cu
’
a c´ac sˆo
´
h˜u
.
uty
’
p
1
, ,p
n
.
II. T´ıch phˆan da
.
ng
R(x,
√
ax
2
+ bx + c)dx, a =0,b
2
− 4ac = 0 (10.9)
38 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
c´o thˆe
’
h˜u
.
uty
’
h´oa nh`o
.
ph´ep thˆe
´
Euler:
(i)
√
ax
2
+ bx + c = ±
√
ax ±t,nˆe
´
u a>0;
(ii)
√
ax
2
+ bx + c = ±xt ±
√
c,nˆe
´
u c>0;
(iii)
√
ax
2
+ bx + c = ±(x −x
1
)t
√
ax
2
+ bx + c = ±(x −x
2
)t
trong d
´o x
1
v`a x
2
l`a c´ac nghiˆe
.
m thu
.
.
c kh´ac nhau cu
’
a tam th´u
.
cbˆa
.
c hai
ax
2
+ nbx + c. (Dˆa
´
uo
.
’
c´ac vˆe
´
pha
’
icu
’
ad
˘a
’
ng th´u
.
c c´o thˆe
’
lˆa
´
y theo tˆo
’
ho
.
.
pt`uy ´y).
III. T´ıch phˆan cu
’
a vi phˆan nhi
.
th´u
.
c. D
´o l `a n h ˜u
.
ng t´ıch phˆan da
.
ng
x
m
(ax
n
+ b)
p
dx (10.10)
trong d
´o a, b ∈ R, m, n, p ∈ Q v`a a =0,b =0,n =0,p = 0; biˆe
’
uth´u
.
c
x
m
(zx
n
+ b)
p
du
.
o
.
.
cgo
.
i l`a vi phˆan nhi
.
th ´u
.
c.
T´ıch phˆan vi phˆan nhi
.
th ´u
.
c (10.10) d
u
.
ad
u
.
o
.
.
cvˆe
`
t´ıch phˆan h`am
h˜u
.
uty
’
trong ba tru
.
`o
.
ng ho
.
.
p sau d
ˆay:
1) p l`a sˆo
´
nguyˆen,
2)
m +1
n
l`a sˆo
´
nguyˆen,
3)
m +1
n
+ p l`a sˆo
´
nguyˆen.
D
-
i
.
nh l´y (Trebu
.
s´ep). T´ıch phˆan vi phˆan nhi
.
th´u
.
c (10.10) biˆe
’
udiˆe
˜
n
d
u
.
o
.
.
cdu
.
´o
.
ida
.
ng h˜u
.
uha
.
n nh`o
.
c´ac h`am so
.
cˆa
´
p(t´u
.
cl`ad
u
.
ad
u
.
o
.
.
cvˆe
`
t´ıch phˆan h`am h˜u
.
uty
’
hay h˜u
.
uty
’
h´oa d
u
.
o
.
.
c) khi v`a chı
’
khi ´ıt nhˆa
´
tmˆo
.
t
trong ba sˆo
´
p,
m +1
n
,
m +1
n
+ p l`a sˆo
´
nguyˆen.
1) Nˆe
´
u p l`a sˆo
´
nguyˆen th`ı ph´ep h˜u
.
uty
’
h´oa s˜e l`a
x = t
N
trong d´o N l`a mˆa
˜
usˆo
´
chung cu
’
a c´ac phˆan th´u
.
c m v`a n.
2) Nˆe
´
u
m +1
n
l`a sˆo
´
nguyˆen th`ı d
˘a
.
t
ax
n
+ b = t
M
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 39
trong d´o M l`a mˆa
˜
usˆo
´
cu
’
a p.
3) Nˆe
´
u
m +1
n
+ p l`a sˆo
´
nguyˆen th`ı d
˘a
.
t
a + bx
−n
= t
M
trong d´o M l`a mˆa
˜
usˆo
´
cu
’
a p.
C
´
AC V
´
IDU
.
V´ı d u
.
1. T´ınh
1) I
1
=
x +
3
√
x
2
+
6
√
x
x(1 +
3
√
x)
dx , 2) I
2
=
dx
3
(2 + x)(2 − x)
5
·
Gia
’
i. 1) T´ıch phˆan d
˜a cho c´o da
.
ng I, trong d´o p
1
=1,p
2
=
1
3
,
p
3
=
1
6
.Mˆa
˜
usˆo
´
chung cu
’
a p
1
,p
2
,p
3
l`a m = 6. Do d´o t a d ˘a
.
t x = t
6
.
Khi d
´o:
I =6
t
6
+ t
4
+ t
t
6
(1 + t
2
)
t
5
dt =6
t
5
+ t
3
+1
1+t
2
dt
=6
t
3
dt +6
dt
1+t
2
=
3
2
3
√
x
2
+ 6arctg
6
√
x + C.
2) B˘a
`
ng ph´ep biˆe
´
nd
ˆo
’
iso
.
cˆa
´
p ta c´o
I
2
=
3
2 − x
2+x
dx
(2 − x)
2
·
D
´o l`a t´ıch phˆan da
.
ng I. Ta d˘a
.
t
2 − x
2+x
= t
3
v`a thu du
.
o
.
.
c
x =2
1 − t
3
1+t
3
,dx= −12
t
2
dt
(1 + t
3
)
2
·
40 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
T`u
.
d
´o
I
2
= −12
t
3
(t
3
+1)
2
dt
16t
6
(t
3
+1)
2
= −
3
4
dt
t
3
=
3
8
3
2+x
2 − x
2
+ C.
V´ı du
.
2. T´ınh c´ac t´ıch phˆan
1) I
1
=
dx
x
√
x
2
+ x +1
, 2) I
2
=
dx
(x − 2)
√
−x
2
+4x − 3
,
3) I
3
=
dx
(x +1)
√
1+x − x
2
, ·
Gia
’
i. 1) T´ıch phˆan I
1
l`a t´ıch phˆan da
.
ng II v`a a =1> 0nˆentasu
.
’
du
.
ng ph´ep thˆe
´
Euler (i)
√
x
2
+ x +1=x + t, x
2
+ x +1=x
2
+2tx + t
2
x =
t
2
− 1
1 − 2t
,
√
x
2
+ x +1=x + t =
−t
2
+ t −1
1 − 2t
dx =
2(−t
2
+ t −1)
(1 − 2t)
2
dt.
T`u
.
d
´o
I
1
=2
dt
t
2
− 1
=ln
1 − t
1+t
+ C =ln
1+x −
√
x
2
+ x +1
1 − x +
√
x
2
+ x +1
+ C.
2) D
ˆo
´
iv´o
.
i t´ıch phˆan I
2
(da
.
ng II) ta c´o
−x
2
+4x − 3=−(x − 1)(x − 3)
v`a do d
´o ta su
.
’
du
.
ng ph´ep thˆe
´
Euler (iii):
√
−x
2
+4x − 3=t(x − 1).
Khi d
´o
−(x − 1)(x − 3) = t
2
(x −1)
2
, −(x −3) = t
2
(x −1),t=
3 − x
x − 1
,
x =
t
2
+3
t
2
+1
,
√
−x
2
+4x − 3=t(x − 1) =
2t
t
2
+1
dx =
−4tdt
(t
2
+1)
2
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 41
v`a thu du
.
o
.
.
c
I
2
=2
dt
t
2
−1
=ln
1 − t
1+t
+ C =ln
√
x − 1 −
√
3 − x
√
x − 1+
√
3 − x
+ C.
3) D
ˆo
´
iv´o
.
i t´ıch phˆan I
3
(da
.
ng III) ta c´o C =1> 0. Ta su
.
’
du
.
ng
ph´ep thˆe
´
Euler (ii) v`a
√
1+x − x
2
= tx − 1, 1+x −x
2
= t
2
x
2
− 2tx +1,
x =
2t +1
t
2
+1
,
√
1+x − x
2
= tx − 1=
t
2
+ t − 1
t
2
+1
,
t =
1+
√
1+x − x
2
x
,dx=
−2(t
2
+ t − 1)
(t
2
+1)
2
·
Do d
´o
I
3
= −2
dt
t
2
+2t +2
= −2
d(t +1)
1+(t +1)
2
= −2arctg(t +1)+C
= −2arctg
1+x +
√
1+x − x
2
x
+ C.
V´ı d u
.
3. T´ınh c´ac t´ıch phˆan
1) I
1
=
√
x
(1 +
3
√
x)
2
dx, x 0; 2) I
2
=
√
x
4
1 −
1
√
x
3
dx;
3) I
3
=
dx
x
2
3
(1 + x
3
)
5
·
Gia
’
i. 1) Ta c´o
I
1
=
x
1
2
1+x
1
3
−2
dx,
trong d
´o m =
1
2
, n =
1
3
, p = −2, mˆa
˜
usˆo
´
chung cu
’
a m v`a n b˘a
`
ng 6.
V`ı p = −2 l`a sˆo
´
nguyˆen, ta ´ap du
.
ng ph´ep d
ˆo
’
ibiˆe
´
n x = t
6
v`a thu du
.
o
.
.
c
I
1
=6
t
8
(1 + t
2
)
2
dt =6
t
4
−2t
2
+3−
4t
2
+3
(1 + t
2
)
2
dt
=
6
5
t
5
−4t
3
+18t −18
dt
1+t
2
− 6
t
2
(1 + t
2
)
2
dt.
42 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
V`ı
t
2
dt
(1 + t
2
)
2
= −
1
2
td
1
1+t
2
= −
t
2(1 + t
2
)
+
1
2
arctgt
nˆen cuˆo
´
ic`ung ta thu d
u
.
o
.
.
c
I
1
=
6
5
x
5/6
− 4x
1/2
+18x
1/6
+
3x
1/6
1+x
1/3
− 21arctgx
1/6
+ C.
2) Ta viˆe
´
t I
2
du
.
´o
.
ida
.
ng
I
2
=
x
1
2
1 − x
−
3
2
1
4
dx.
O
.
’
d
ˆay m =
1
2
, n = −
3
2
, p =
1
4
v`a
m +1
n
= −1 l`a sˆo
´
nguyˆen v`a ta
c´o tru
.
`o
.
ng ho
.
.
pth´u
.
hai. Ta su
.
’
du
.
ng ph´ep d
ˆo
’
ibiˆe
´
n
1 −
1
√
x
3
= t
4
.
Khi d
´o x =(1−t
4
)
−
2
3
, dx =
8
3
(1 − t
4
)
−
5
3
t
3
dt v`a do vˆa
.
y
I
2
=
8
3
t
4
(1 − t
4
)
2
dt =
2
3
td
1
1 − t
4
=
2
3
t
1 − t
4
−
dt
1 − t
2
=
2t
3(1 − t
4
)
−
1
3
1
1 − t
2
+
1
1+t
2
dt
=
2t
3(1 − t
4
)
−
1
6
ln
1+t
1 − t
−
1
3
arctgt + C,
trong d
´o t =
1 − x
−3/2
1/4
.
3) Ta viˆe
´
t I
3
du
.
´o
.
ida
.
ng
I
3
=
x
−2
(1 + x
3
)
−
5
3
dx.
O
.
’
d
ˆay m = −2, n =3,p = −
5
3
v`a
m +1
n
+ p = −2 l`a sˆo
´
nguyˆen.
Do vˆa
.
y ta c´o tru
.
`o
.
ng ho
.
.
pth´u
.
ba. Ta thu
.
.
chiˆe
.
n ph´ep d
ˆo
’
ibiˆe
´
n
1+x
−3
= t
3
⇒ 1+x
3
= t
3
x
3
.
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 43
T`u
.
d
´o
x
3
=
1
t
3
− 1
, 1+x
3
=
t
3
t
3
− 1
,x=(t
3
− 1)
−
1
3
dx = −t
2
(t
3
− 1)
−
4
3
dt, x
−2
=(t
3
− 1)
2
3
.
Do vˆa
.
y
I
3
= −
(t
3
−1)
2/3
t
3
t
3
− 1
−5/3
t
2
(t
3
− 1)
−
4
3
dt =
1 − t
3
t
3
dt
=
t
−3
dt −
dt =
t
−2
−2
−t + C = C −
1+2t
3
2t
3
= C −
2+3x
3
2x
3
(1 + x
3
)
2
·
44 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
B
`
AI T
ˆ
A
.
P
T´ınh c´ac t´ıch phˆan (1-15)
1.
dx
√
2x − 1 −
3
√
2x − 1
.
(D
S. u
3
+
3
2
u
2
+3u +3ln|u − 1|,u
6
=2x − 1)
2.
xdx
(3x − 1)
√
3x − 1
.
(D
S.
2
9
3x − 2
√
3x − 1
)
3.
1 − x
1+x
dx
x
.
(D
S.
1 −
√
1 − x
2
x
−arc sin x)
4.
3
x +1
x − 1
dx
x +1
.
(D
S. −
1
2
ln
(1 − t)
2
1+t + t
2
+
√
3arctg
2t +1
√
3
,t=
3
x +1
x − 1
)
5.
√
x +1−
√
x − 1
√
x +1+
√
x − 1
dx.
(D
S.
1
2
(x
2
−x
√
x
2
− 1+ln|x +
√
x
2
− 1|)
6.
xdx
√
x +1−
3
√
x +1
.
(D
S. 6
1
9
u
9
+
1
8
u
8
+
1
7
u
7
+
1
6
u
6
+
1
5
u
5
+
1
4
u
4
,u
6
= x +1)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 45
7.
(x − 2)
1+x
1 − x
dx.
(D
S.
1 −
1
2
x
√
1 − x
2
−
3
2
arc sin x)
8.
3
x +1
x −1
dx
(x − 1)
3
.
(D
S.
3
16
3
x +1
x − 1
4
−
3
28
3
x +1
x − 1
3
)
9.
dx
(x − 1)
3
(x − 2)
.(D
S. 2
x − 2
x − 1
)
Chı
’
dˆa
˜
n. Viˆe
´
t
(x − 1)
3
(x − 2) = (x − 1)(x − 2)
x − 1
x − 2
,d
˘a
.
t
t =
x − 1
x − 2
.
10.
dx
3
(x −1)
2
(x +1)
.
(D
S.
1
2
ln
u
2
+ u +1
u
2
− 2u +1
−
√
3arctg
2u +1
√
3
,u
3
=
x +1
x − 1
)
11.
dx
3
(x +1)
2
(x − 1)
4
.(DS.
3
2
3
1+x
x − 1
)
12.
dx
4
(x −1)
3
(x +2)
5
.(DS.
4
3
4
x − 1
x +2
)
13.
dx
3
(x −1)
7
(x +1)
2
.(DS.
3
16
3x − 5
x − 1
3
x +1
x − 1
)
14.
dx
6
(x −7)
7
(x − 5)
5
.(DS. −3
6
x − 5
x − 7
)
15.
dx
n
(x −a)
n+1
(x − b)
n−1
, a = b.(DS.
n
b − a
n
x − b
x − a
)
46 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
16.
√
x +1−
√
x − 1
√
x +1+
√
x − 1
dx.
(D
S.
x
2
3
−
x
√
x
2
− 1
2
+
1
2
ln |x +
√
x
2
−1|)
Su
.
’
du
.
ng c´ac ph´ep thˆe
´
Euler d
ˆe
’
t´ınh c´ac t´ıch phˆan sau dˆay (17-22)
17.
dx
x
√
x
2
+ x +1
.(D
S. ln
1+x −
√
x
2
+ x +1
1 − x +
√
x
2
+ x +1
)
18.
dx
(x − 2)
√
−x
2
+4x − 3
.(D
S. ln
√
x − 1 −
√
3 − x
√
x − 1+
√
3 − x
)
19.
dx
(x +1)
√
1+x − x
2
.(DS. −2arctg
1+x +
√
1+x − x
2
x
)
20.
dx
(x − 1)
√
x
2
+ x +1
.
(D
S.
√
3
3
ln
x − 1
3+3x +2
3(x
2
+ x +1)
)
21.
(x −1)dx
(x
2
+2x)
√
x
2
+2x
.(D
S.
1+2x
√
x
2
+2x
)
22.
5x +4
√
x
2
+2x +5
dx.
(D
S. 5
√
x
2
+2x +5− ln
x +1+
√
x
2
+2x +5
)
Chı
’
dˆa
˜
n. C´o thˆe
’
d
ˆo
’
ibiˆe
´
n t =
1
2
(x
2
+2x +5)
= x +1.
T´ınh c´ac t´ıch phˆan cu
’
a vi phˆan nhi
.
th ´u
.
c
23.
x
−
1
3
(1 − x
1/6
)
−1
dx.(DS. 6x
1
6
+3x
1
3
+2x
1
2
+6ln
x
1
6
− 1
)
24.
x
−
2
3
(1 + x
1
3
)
−3
dx.(DS. −
3
2
(1 + x
1
3
)
−2
)
25.
x
−
1
2
(1 + x
1
4
)
−10
dx.(DS.
4
9
(1 + x
1
4
)
−9
−
1
2
(1 + x
1
4
)
−8
)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 47
26.
x
1+
3
√
x
2
dx.(DS. 3
t
5
5
−
2t
3
3
+ t
, t =
√
1+x
2/3
)
27.
x
3
(1 + 2x
2
)
−
2
3
dx.(DS.
x
2
+1
2
√
2x
2
+1
)
28.
dx
x
4
√
1+x
2
.(DS.
1
3
x
−3
(2x
2
− 1)
√
x
2
+1)
29.
dx
x
2
(1 + x
3
)
5/3
.(DS. −
1
8
x
−1
(3x + 4)(2 + x
3
)
−
2
3
)
30.
dx
√
x
3
3
1+
4
√
x
3
.(DS. −2
3
(x
−
3
4
+1)
2
)
31.
dx
3
√
x
2
(
3
√
x +1)
3
.(DS. −
3
2(
3
√
x +1)
2
)
32.
3
√
x
3
√
x +1
dx.
(D
S. 6
u
7
7
−
3
5
u
5
+ u
3
−u
2
,u
2
=
3
√
x +1
)
33.
dx
x
6
√
x
2
− 1
.
(D
S.
u
5
5
−
2u
3
3
+ u, u =
√
1 − x
−2
)
34.
dx
x
3
√
1+x
5
.
(D
S.
1
10
ln
u
2
−2u +1
u
2
+ u +1
+
√
3
5
arctg
2u +1
√
3
,u
3
=1+x
5
)
35.
x
7
√
1+x
2
dx.
(D
S.
u
9
9
−
3u
7
7
+
3u
5
5
−
u
3
3
,u
2
=1+x
2
)
48 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
36.
dx
3
√
1+x
3
.
(D
S.
1
6
ln
u
2
+ u +1
u
2
− 2u +1
−
1
√
3
arctg
2u +1
√
3
,u
3
=1+x
−3
)
37.
dx
4
√
1+x
4
.
(D
S.
1
4
ln
u +1
u − 1
−
1
2
arctgu, u
4
=1+x
−4
)
38.
3
√
x − x
3
dx.
(D
S.
u
2(u
3
+1)
−
1
12
ln
u
2
+2u +1
u
2
− u +1
−
1
2
√
3
arctg
2u − 1
√
3
,u
3
= x
−2
− 1)
10.2.3 T´ıch phˆan c´ac h`am lu
.
o
.
.
ng gi´ac
I. T´ıch phˆan da
.
ng
R(sin x, cos x)dx (10.11)
trong d
´o R(u, v) l`a h`am h˜u
.
uty
’
cu
’
a c´ac biˆe
´
n u b`a v luˆon luˆon c´o thˆe
’
h˜u
.
uty
’
h´oa d
u
.
o
.
.
c nh`o
.
ph´ep d
ˆo
’
ibiˆe
´
n t =tg
x
2
, x ∈ (−π,π). T`u
.
d
´o
sin x =
2t
1+t
2
, cos x =
1 − t
2
1+t
2
,dx=
2dt
1+t
2
·
Nhu
.
o
.
.
cd
iˆe
’
mcu
’
aph´ep h˜u
.
uty
’
h´oa n`ay l`a n´o thu
.
`o
.
ng d
u
.
ad
ˆe
´
nnh˜u
.
ng
t´ınh to´an rˆa
´
tph´u
.
cta
.
p.
V`ıvˆa
.
y, trong nhiˆe
`
u tru
.
`o
.
ng ho
.
.
p ph´ep h˜u
.
uty
’
h´oa c´o thˆe
’
thu
.
.
chiˆe
.
n
d
u
.
o
.
.
c nh`o
.
nh˜u
.
ng ph´ep d
ˆo
’
ibiˆe
´
n kh´ac.
II. Nˆe
´
u R(−sin x, cos x)=−R(sin x, cos x) th`ı su
.
’
du
.
ng ph´ep d
ˆo
’
i
biˆe
´
n
t = cos x, x ∈ (0,π)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 49
v`a l ´uc d´o
dx = −
dt
√
1 − t
2
III. Nˆe
´
u R(sin x, −cos x)=−R(sin x, cos x) th`ı su
.
’
du
.
ng ph´ep d
ˆo
’
i
biˆe
´
n
t = sin x, dx =
dt
√
1 − t
2
,x∈
−
π
2
,
π
2
.
IV. Nˆe
´
u R(−sin x, −cos x)=R(sin x, cos x) th`ı ph´ep h˜u
.
uty
’
h´oa
s˜e l`a t =tgx, x ∈
−
π
2
,
π
2
:
sin x =
t
√
1+t
2
, cos x =
1
√
1+t
2
,x= arctgt, dx =
dt
1+t
2
·
V. Tru
.
`o
.
ng ho
.
.
p riˆeng cu
’
a t´ıch phˆan da
.
ng (10.11) l`a t´ıch phˆan
sin
m
x cos
n
xdx, m, n ∈ Z (10.12)
(i) Nˆe
´
usˆo
´
m le
’
th`ı d
˘a
.
t t = cos x,nˆe
´
u n le
’
th`ı d˘a
.
t sin x = t.
(ii) Nˆe
´
u m v`a n l`a nh˜u
.
ng sˆo
´
ch˘a
˜
n khˆong ˆam th`ı tˆo
´
tho
.
nhˆe
´
t l`a thay
sin
2
x v`a cos
2
x theo c´ac cˆong th´u
.
c
sin
2
x =
1
2
(1 − cos 2x), cos
2
x =
1
2
(1 + cos 2x).
(iii) Nˆe
´
u m v`a n ch˘a
˜
n, trong d
´o c´o mˆo
.
tsˆo
´
ˆam th`ı ph´ep dˆo
’
ibiˆe
´
ns˜e
l`a tgx = t hay cotgx = t.
(iv) Nˆe
´
u m + n = −2k, k ∈ N th`ı viˆe
´
tbiˆe
’
uth´u
.
cdu
.
´o
.
idˆa
´
ut´ıch
phˆan bo
.
’
ida
.
ng phˆan th´u
.
c v`a t´ach cos
2
x (ho˘a
.
c sin
2
x) ra kho
’
imˆa
˜
usˆo
´
.
Biˆe
’
uth´u
.
c
dx
cos
2
x
(ho˘a
.
c
dx
sin
2
x
)d
u
.
o
.
.
cthaybo
.
’
i d(tgx) (ho˘a
.
c d(cotgx))
v`a ´ap du
.
ng ph´ep d
ˆo
’
ibiˆe
´
n t =tgx (ho˘a
.
c t = cotgx).
VI. T´ıch phˆan da
.
ng
sin
α
x cos
β
xdx, α, β ∈ Q. (10.13)
50 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
B˘a
`
ng ph´ep dˆo
’
ibiˆe
´
n sin
2
x = t ta thu du
.
o
.
.
c
I =
1
2
t
α−1
2
(1 − t)
β−1
2
dt
v`a b`ai to´an d
u
.
o
.
.
cquyvˆe
`
t´ıch phˆan cu
’
a vi phˆan nhi
.
th ´u
.
c.
C
´
AC V
´
IDU
.
V´ı du
.
1. T´ınh t´ıch phˆan
I =
dx
3 sin x + 4 cos x +5
Gia
’
i. D
˘a
.
t t =tg
x
2
, x ∈ (−π, π). Khi d
´o
I =2
dt
t
2
+6t +9
=2
(t +3)
−2
dt
= −
2
t +3
+ C = −
2
3+tg
x
2
+ C.
V´ı du
.
2. T´ınh
J =
dx
(3 + cos 5x)sin5x
Gia
’
i. D
˘a
.
t5x = t.Tathudu
.
o
.
.
c
J =
1
5
dt
(3 + cos t) sin t
v`a (tru
.
`o
.
ng ho
.
.
p II) do d
´ob˘a
`
ng c´ach d˘a
.
tph´ep dˆo
’
ibiˆe
´
n z = cos t ta c´o
J =
1
5
dz
(z + 3)(z
2
− 1)
=
1
5
A
z − 1
+
B
z − 1
+
C
z +3
dz
=
1
5
1
8(z − 1)
−
1
4(z +1)
+
1
8(z +3)
dz
=
1
5
1
8
ln |z − 1|−
1
4
ln |z +1| +
1
8
ln |z +3|
+ C
=
1
40
ln
(z − 1)( z +3)
(z +1)
2
+ C
=
1
40
ln
cos
2
x + 2 cos 5x − 3
(cos 5x +1)
2
+ C.
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 51
V´ı d u
.
3. T´ınh
J =
2 sin x + 3 cos x
sin
2
x cos x + 9 cos
3
x
dx
Gia
’
i. H`am du
.
´o
.
idˆa
´
u t´ıch phˆan c´o t´ınh chˆa
´
tl`a
R(−sin x, −cos x)=R(sin x, cos x).
Do d
´o ta su
.
’
du
.
ng ph´ep d
ˆo
’
ibiˆe
´
n t =tgx, x ∈
−
π
2
,
π
2
. Chia tu
.
’
sˆo
´
v`a mˆa
˜
usˆo
´
cu
’
abiˆe
’
uth´u
.
cdu
.
´o
.
idˆa
´
u t´ıch phˆan cho cos
3
x ta c´o
J =
2tgx +3
tg
2
x +9
d(tg x)=
2t +3
t
2
+9
dt
= ln(t
2
+ 9) + arctg
t
3
+ C
= ln(tg
2
x + 9) + arctg
tgx
3
+ C.
V´ı d u
.
4. T´ınh
J =
dx
sin
6
x + cos
6
x
Gia
’
i.
´
Ap du
.
ng cˆong th ´u
.
c
cos
2
x =
1
2
(1 + cos 2x), sin
2
x =
1
2
(1 − sin 2x)
ta thu d
u
.
o
.
.
c
cos
6
x + sin
6
x =
1
4
(1 + 3 cos
2
2x).
D
˘a
.
t t = tg2x,tat`ımdu
.
o
.
.
c
J =
4dx
1 + 3 cos
2
2x
=2
dt
t
2
+4
= arctg
t
2
+ C = arctg
tg2x
2
+ C.
52 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
V´ı du
.
5. T´ınh
J =
sin
3
2
x cos
1
2
xdx.
Gia
’
i. D
˘a
.
t z = sin
2
x ta thu du
.
o
.
.
c
J =
1
2
z
1/4
(1 − z)
−
1
4
dx.
D
´o l`a t´ıch phˆan cu
’
a vi phˆan nhi
.
th ´u
.
cv`a
m +1
n
+ p =
1
4
+1
1
−
1
4
=1.
Do vˆa
.
y ta thu
.
.
chiˆe
.
n ph´ep d
ˆo
’
ibiˆe
´
n
1
z
− 1=t
4
, −
dz
z
2
=4t
3
dt, z
2
=
1
(t
4
+1)
2
v`a do d´o
J = −2
t
2
(t
4
+1)
2
dt.
D
˘a
.
t t =
1
y
ta thu d
u
.
o
.
.
c
J =2
y
4
(1 + y
4
)
2
dy.
Thu
.
.
chiˆe
.
n ph´ep t´ıch phˆan t`u
.
ng phˆa
`
nb˘a
`
ng c´ach d
˘a
.
t
u = y, dv =
y
3
(1 + y
4
)
2
dy ⇒ du = dy, v = −
1
4(1 + y
2
)
ta thu d
u
.
o
.
.
c
J =2
−
y
4(1 + y
4
)
+
1
4
dy
1+y
4
= −
y
2(1 + y
4
)
+
1
2
J
1
.
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 53
Dˆe
’
t´ınh J
1
ta biˆe
’
udiˆe
˜
ntu
.
’
sˆo
´
cu
’
abiˆe
’
uth´u
.
cdu
.
´o
.
idˆa
´
ut´ıch phˆan
nhu
.
sau:
1=
1
2
(y
2
+1)−(y
2
− 1)
v`a khi d
´o
J
1
=
1
2
y
2
+1
y
4
+1
dy −
1
2
y
2
−1
y
4
+1
dy
=
1
2
1+
1
y
2
dy
y
2
+
1
y
2
−
1
2
1 −
1
y
2
dy
y
2
+
1
y
2
=
1
2
d
y +
1
y
y −
1
y
2
+2
−
1
2
d
y +
1
y
y +
1
y
2
−2
=
1
2
√
2
arctg
y −
1
y
√
2
−
1
4
√
2
ln
y +
1
y
−
√
2
yb +
1
y
+
√
2
+ C.
Cuˆo
´
ic`ung ta thu d
u
.
o
.
.
c
J = −
y
2(1 + y
4
)
+
1
4
√
2
arctg
y −
1
4
√
2
−
1
8
√
2
ln
y +
1
y
−
√
2
y +
1
y
+
√
2
+ C
trong d
´o
y =
1
t
,t=
4
1
z
− 1 ,z= sin
2
x.
B
`
AI T
ˆ
A
.
P
T´ınh c´ac t´ıch phˆan b˘a
`
ng c´ach su
.
’
du
.
ng c´ac cˆong th´u
.
clu
.
o
.
.
ng gi´ac
d
ˆe
’
biˆe
´
ndˆo
’
i h`am du
.
´o
.
idˆa
´
u t´ıch phˆan.
54 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
1.
sin
3
xdx.(DS. −cos x +
cos
3
x
3
)
2.
cos
4
xdx.(DS.
3x
8
+
sin 2x
4
+
sin 4x
32
)
3.
sin
5
xdx.(DS.
2
3
cos
3
x −
cos
5
x
5
− cos x)
4.
cos
7
xdx.(DS. sin x −sin
3
x +
3 sin
5
x
5
−
sin
7
x
7
)
5.
cos
2
x sin
2
xdx.(DS.
x
8
−
sin 4x
32
)
6.
sin
3
x cos
2
xdx.(DS.
cos
5
x
5
−
cos
3
x
3
)
7.
cos
3
x sin
5
xdx.(DS.
sin
6
x
6
−
sin
8
x
8
)
8.
dx
sin 2x
.(D
S.
1
2
ln |tgx|)
9.
dx
cos
x
3
.(D
S. 3 ln
tg
π
4
+
x
6
)
10.
sin x + cos x
sin 2x
dx.(D
S.
1
2
ln
tg
x
2
+ln
tg
π
4
+
x
2
)
11.
sin
2
x
cos
6
x
dx.(D
S.
tg
5
x
5
+
tg
3
x
3
)
Chı
’
dˆa
˜
n. D
˘a
.
t t =tgx.
12.
sin 3x cos xdx.(D
S. −
1
8
(cos 4x + 2 cos 2x))
13.
sin
x
3
cos
2x
3
dx.(D
S.
3
2
cos
x
3
−
1
2
cos x)
14.
cos
3
x
sin
2
x
dx.(D
S. −
1
sin x
− sin x)
15.
sin
3
x
cos
2
x
dx.(D
S.
1
cos x
+ cos x)
10.2. C´ac l´o
.
p h`am kha
’
t´ıch trong l´o
.
p c´ac h`am so
.
cˆa
´
p 55
16.
cos
3
x
sin
5
x
dx.(D
S. −
cotg
4
x
4
)
17.
sin
5
x
cos
3
x
dx.(D
S.
1
2 cos
2
x
+2ln|cos x|−
cos
2
x
2
)
18.
tg
5
xdx.(DS.
tg
4
x
4
−
tg
2
x
2
− ln |cos x|)
Trong c´ac b`ai to´an sau d
ˆay h˜ay ´ap du
.
ng ph´ep dˆo
’
ibiˆe
´
n
t =tg
x
2
, sin x =
2t
1+t
2
, cos x =
1 − t
2
1+t
2
,x= 2arctgt, dx =
2dt
1+t
2
19.
dx
3 + 5 cos x
.(D
S.
1
4
ln
2+tg
x
2
2 − tg
x
2
)
20.
dx
sin x + cos x
.(D
S.
1
√
2
ln
tg
x
2
+
π
8
)
21.
3 sin x + 2 cos x
2 sin x + 3 cos x
dx.
(D
S.
1
13
(12x − 5ln|2tgx +3|−5ln|cos x|)
22.
dx
1 + sin x + cos x
.(D
S. ln
1+tg
x
2
)
23.
dx
(2 − sin x)(3 − sin x)
.
(D
S.
2
√
3
arctg
2tg
x
2
− 1
√
3
−
1
√
2
arctg
3tg
x
2
− 1
2
√
2
)
T´ınh c´ac t´ıch phˆan da
.
ng
sin
m
x cos
n
xdx, m, n ∈ N.
24.
sin
3
x cos
5
xdx.(DS.
1
8
cos
8
x −
1
6
cos
6
x)
56 Chu
.
o
.
ng 10. T´ıch phˆan bˆa
´
td
i
.
nh
25.
sin
2
x cos
4
xdx.(DS.
1
16
x −
1
4
sin 4x +
1
3
sin
2
2x
)
26.
sin
4
x cos
6
xdx.
(D
S.
1
2
11
sin 8x −
1
2
8
sin 4x +
1
5 · 2
6
sin
5
2x +
3
2
8
x)
27.
sin
4
x cos
2
xdx.(DS.
x
16
−
sin 4x
64
−
sin
2
2x
48
)
28.
sin
4
x cos
5
xdx.(DS.
1
5
sin
5
x −
2
7
sin
7
x +
1
9
sin
9
x)
29.
sin
6
x cos
3
xdx.(DS.
1
7
sin
7
x −
1
9
sin
9
x)
T´ınh c´ac t´ıch phˆan da
.
ng
sin
α
x cos
β
xdx, α, β ∈ Q.
30.
sin
3
x
cos x
3
√
cos x
dx.(D
S.
3
5
cos x
3
√
cos
2
x +
3
3
√
cos x
)
Chı
’
dˆa
˜
n. D
˘a
.
t t = cos x.
31.
dx
3
√
sin
11
x cos x
.(D
S. −
3(1 + 4tg
2
x)
8tg
2
x ·
3
tg
2
x
)
Chı
’
dˆa
˜
n. D
˘a
.
t t =tgx.
32.
sin
3
x
3
√
cos
2
x
dx.(D
S. 3
3
√
cos x
1
7
cos
2
x − 1
)
33.
3
√
cos
2
x sin
3
xdx.(DS. −
3
5
cos
5/3
x +
3
11
cos
11
3
x)
34.
dx
4
√
sin
3
x cos
5
x
.(D
S. 4
4
√
tgx)
35.
sin
3
x
5
√
cos x
dx.(D
S.
5
14
cos
14
5
x −
5
4
cos
4
5
x)
Chu
.
o
.
ng 11
T´ıch phˆan x´ac d
i
.
nh Riemann
11.1 H`am kha
’
t´ıch Riemann v`a t´ıch phˆan x´ac
d
i
.
nh 58
11.1.1 D
-
i
.
nhngh˜ıa 58
11.1.2 D
-
iˆe
`
ukiˆe
.
nd
ˆe
’
h`am kha
’
t´ıch 59
11.1.3 C´ac t´ınh chˆa
´
tco
.
ba
’
ncu
’
at´ıch phˆan x´ac di
.
nh 59
11.2 Phu
.
o
.
ng ph´ap t´ınh t´ıch phˆan x´ac di
.
nh . . . 61
11.3 Mˆo
.
tsˆo
´
´u
.
ng du
.
ng cu
’
at´ıch phˆan x´ac d i
.
nh . 78
11.3.1 Diˆe
.
n t´ıch h`ınh ph˘a
’
ng v`a thˆe
’
t´ıch vˆa
.
tthˆe
’
78
11.3.2 T´ınh d
ˆo
.
d`ai cung v`a diˆe
.
n t´ıch m˘a
.
t tr`on xoay 89
11.4 T´ıch phˆan suy rˆo
.
ng 98
11.4.1 T´ıch phˆan suy rˆo
.
ng cˆa
.
nvˆoha
.
n 98
11.4.2 T´ıch phˆan suy rˆo
.
ng cu
’
a h`am khˆong bi
.
ch˘a
.
n 107
