Bài tập toán cao cấp có lời giải chi tiết

˜
’ THANH
ˆ N THUY
NGUYE

` TA
ˆP
BAI
.
´ CAO CA
ˆ´P
TOAN
Tˆa.p 3
Ph´ep t´ınh t´ıch phˆan. L´
y thuyˆe´t chuˆ˜o i.
Phu.o.ng tr`ınh vi phˆan

’ N DAI HOC QUO
` XUA
ˆ´T BA
ˆ´C GIA HA
` NO
ˆ. I
NHA
.
.


Mu.c lu.c
10 T´ıch phˆ
an bˆ

a´t di.nh
10.1 C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan . . . . .
10.1.1 Nguyˆen h`am v`a t´ıch phˆan bˆa´t di.nh
10.1.2 Phu.o.ng ph´ap dˆo’i biˆe´n . . . . . . .
`an
u.ng phˆ
10.1.3 Phu.o.ng ph´ap t´ıch phˆan t`

. . . . . . .

4
4

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4

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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 . . . .
10.2.1 T´ıch phˆan c´ac h`am h˜
u.u ty’ . . . . . . . . . . . .
10.2.2 T´ıch phˆan mˆo.t sˆo´ h`am vˆo ty’ do.n gia’n . . . . .
10.2.3 T´ıch phˆan c´ac h`am lu.o..ng gi´ac . . . . . . . . . .

12
21

11 T´ıch phˆ
an x´

ac di.nh Riemann
11.1 H`am kha’ t´ıch Riemann v`a t´ıch phˆan x´ac d i.nh . . .
- i.nh ngh˜ıa . . . . . . . . . . . . . . . . . .
11.1.1 D
- iˆ
`eu kiˆe.n dˆe’ h`am kha’ t´ıch . . . . . . . . . .
11.1.2 D
11.1.3 C´ac t´ınh chˆa´t co. ba’n cu’a t´ıch phˆan x´ac di.nh
11.2 Phu.o.ng ph´ap t´ınh t´ıch phˆan x´ac d i.nh . . . . . . .
11.3 Mˆo.t sˆo´ u
´.ng du.ng cu’a t´ıch phˆan x´ac d i.nh . . . . . .
11.3.1 Diˆe.n t´ıch h`ınh ph˘a’ng v`a thˆe’ t´ıch vˆa.t thˆe’ . .

30
30
37
48
57

. .

58

. .
. .

58
59

. .

59

. .
. .

61
78

. .

78

11.3.2 T´ınh dˆo. d`ai cung v`a diˆe.n t´ıch m˘a.t tr`on xoay . .
11.4 T´ıch phˆan suy rˆo.ng . . . . . . . . . . . . . . . . . . . .

89
98

11.4.1 T´ıch phˆan suy rˆo.ng cˆa.n vˆo ha.n . . . . . . . . . 98
11.4.2 T´ıch phˆan suy rˆo.ng cu’a h`am khˆong bi. ch˘a.n . . 107


2

MU
. C LU
.C
`eu biˆ
12 T´ıch phˆ

an h`
am nhiˆ
e´n
12.1 T´ıch phˆan 2-l´o.p . . . . . . . . . . . . . .
`en ch˜
u. nhˆa.t . . .
12.1.1 Tru.`o.ng ho..p miˆ
`en cong . . . . . .
12.1.2 Tru.`o.ng ho..p miˆ
12.1.3 Mˆo.t v`ai u
´.ng du.ng trong h`ınh ho.c
12.2 T´ıch phˆan 3-l´o.p . . . . . . . . . . . . . .
`en h`ınh hˆo.p . . .
12.2.1 Tru.`o.ng ho..p miˆ
`en cong . . . . . .
12.2.2 Tru.`o.ng ho..p miˆ
12.2.3
. . . . . . . . . . . . . . . . . .
12.2.4 Nhˆa.n x´et chung . . . . . . . . . .
12.3 T´ıch phˆan d u.`o.ng . . . . . . . . . . . . .
12.3.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . .
12.3.2 T´ınh t´ıch phˆan du.`o.ng . . . . . .
12.4 T´ıch phˆan m˘a.t . . . . . . . . . . . . . .
12.4.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . .
12.4.2 Phu.o.ng ph´ap t´ınh t´ıch phˆan m˘a.t
12.4.3 Cˆong th´
u.c Gauss-Ostrogradski .
12.4.4 Cˆong th´
u.c Stokes . . . . . . . . .

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117
118
118
118
121
133
133
134
136
136
144
144

146
158
158
160
162
162

˜i
13 L´
y thuyˆ
e´t chuˆ
o
13.1 Chuˆ˜o i sˆo´ du.o.ng . . . . . . . . . . . . . . . . . . . . . .
13.1.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . . .
13.1.2 Chuˆo˜ i sˆo´ du.o.ng . . . . . . . . . . . . . . . . . .
13.2 Chuˆ˜o i hˆo.i tu. tuyˆe.t d ˆo´i v`a hˆo.i tu. khˆong tuyˆe.t d ˆo´i . . .
13.2.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . . .
13.2.2 Chuˆo˜ i dan dˆa´u v`a dˆa´u hiˆe.u Leibnitz . . . . . .
13.3 Chuˆ˜o i l˜
uy th`
u.a . . . . . . . . . . . . . . . . . . . . . .
13.3.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . . .
- iˆ
`eu kiˆe.n khai triˆe’n v`a phu.o.ng ph´ap khai triˆe’n
13.3.2 D
13.4 Chuˆo˜ i Fourier . . . . . . . . . . . . . . . . . . . . . . .
13.4.1 C´ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . . .

177
178

178
179
191
191
192
199
199
201
211
211

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MU
. C LU
.C

3

`e su.. hˆo.i tu. cu’a chuˆ˜o i Fourier . . . 212
13.4.2 Dˆa´u hiˆe.u du’ vˆ
14 Phu.o.ng tr`ınh vi phˆ
an
224
14.1 Phu.o.ng tr`ınh vi phˆan cˆa´p 1 . . . . . . . . . . . . . . . 225

14.1.1 Phu.o.ng tr`ınh t´ach biˆe´n . . . . . . . . . . . . . . 226
14.1.2 Phu.o.ng tr`ınh d ˘a’ng cˆa´p . . . . . . . . . . . . . 231
14.1.3 Phu.o.ng tr`ınh tuyˆe´n t´ınh . . . . . . . . . . . . . 237
14.1.4 Phu.o.ng tr`ınh Bernoulli . . . . . . . . . . . . . . 244
`an . . . . . . . . 247
14.1.5 Phu.o.ng tr`ınh vi phˆan to`an phˆ
14.1.6 Phu.o.ng tr`ınh Lagrange v`a phu.o.ng tr`ınh Clairaut255
14.2 Phu.o.ng tr`ınh vi phˆan cˆa´p cao . . . . . . . . . . . . . . 259
14.2.1 C´ac phu.o.ng tr`ınh cho ph´ep ha. thˆa´p cˆa´p . . . . 260
14.2.2 Phu.o.ng tr`ınh vi phˆan tuyˆe´n t´ınh cˆa´p 2 v´o.i hˆe.
sˆo´ h˘a`ng . . . . . . . . . . . . . . . . . . . . . . 264
`an nhˆa´t
14.2.3 Phu.o.ng tr`ınh vi phˆan tuyˆe´n t´ınh thuˆ
cˆa´p n (ptvptn cˆa´p n ) v´o.i hˆe. sˆo´ h˘a`ng . . . . . . 273
14.3 Hˆe. phu.o.ng tr`ınh vi phˆan tuyˆe´n t´ınh cˆa´p 1 v´o.i hˆe. sˆo´ h˘a`ng290
`e phu.o.ng tr`ınh vi phˆ
15 Kh´
ai niˆ
e.m vˆ
an da.o h`
am riˆ
eng
15.1 Phu.o.ng tr`ınh vi phˆan cˆa´p 1 tuyˆe´n t´ınh dˆo´i v´o.i c´ac da.o
h`am riˆeng . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Gia’i phu.o.ng tr`ınh d a.o h`am riˆeng cˆa´p 2 d o.n gia’n nhˆa´t
y to´an co. ba’n . . . . . . . . . .
15.3 C´ac phu.o.ng tr`ınh vˆa.t l´
`en s´ong . . . . . . . . . . . .
15.3.1 Phu.o.ng tr`ınh truyˆ
.

.
`en nhiˆe.t . . . . . . . . . . . .
15.3.2 Phu o ng tr`ınh truyˆ
15.3.3 Phu.o.ng tr`ınh Laplace . . . . . . . . . . . . . .
T`
ai liˆ
e.u tham kha’o . . . . . . . . . . . . . . . . . . . . .

304
306
310
313
314
317
320
327


Chu.o.ng 10
T´ıch phˆ
an bˆ
a´t di.nh

ap t´ınh t´ıch phˆ
an . . . . . .
10.1 C´
ac phu.o.ng ph´

4

10.1.1 Nguyˆen h`
am v`
a t´ıch phˆ
an bˆ
a´t di.nh . . . . .

4

10.1.2 Phu.o.ng ph´
ap dˆ
o’i biˆe´n . . . . . . . . . . . . 12
`an . . . . . 21
10.1.3 Phu.o.ng ph´
ap t´ıch phˆ
an t`
u.ng phˆ
10.2 C´
ac l´
o.p h`
am kha’ t´ıch trong l´
o.p c´
ac h`
am
.
so cˆ
a´p . . . . . . . . . . . . . . . . . . . . . . 30
10.2.1 T´ıch phˆ
an c´
ac h`
am h˜

u.u ty’ . . . . . . . . . 30
10.2.2 T´ıch phˆ
an mˆ
o.t sˆ
o´ h`
am vˆ
o ty’ do.n gia’n . . . 37
10.2.3 T´ıch phˆ
an c´
ac h`
am lu.o..ng gi´
ac . . . . . . . 48

10.1

ap t´ınh t´ıch phˆ
an
C´
ac phu.o.ng ph´

10.1.1

Nguyˆ
en h`
am v`
a t´ıch phˆ
an bˆ
a´t di.nh

- i.nh ngh˜ıa 10.1.1. H`am F (x) du.o..c go.i l`a nguyˆen h`am cu’a h`am

D
f (x) trˆen khoa’ng n`ao d´o nˆe´u F (x) liˆen tu.c trˆen khoa’ng d´o v`a kha’ vi


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

5

ta.i mˆo˜ i diˆe’m trong cu’a khoa’ng v`a F (x) = f (x).
- i.nh l´
`on ta.i nguyˆen h`am) Mo.i h`
`e su.. tˆ
am liˆen tu.c trˆen
D
y 10.1.1. (vˆ
`eu c´
doa.n [a, b] dˆ
o nguyˆen h`
am trˆen khoa’ng (a, b).
- i.nh l´
D
y 10.1.2. C´
ac nguyˆen h`
am bˆ
a´t k`y cu’a c`
ung mˆ
o.t h`
am l`
a chı’
.

`ng sˆ
kh´
ac nhau bo’ i mˆ
o.t h˘
a
o´ cˆ
o.ng.
Kh´ac v´o.i da.o h`am, nguyˆen h`am cu’a h`am so. cˆa´p khˆong pha’i bao
2
ung l`a h`am so. cˆa´p. Ch˘a’ng ha.n, nguyˆen h`am cu’a c´ac h`am e−x ,
gi`o. c˜
1 cos x sin x
,
,
,... l`a nh˜
u.ng h`am khˆong so. cˆa´p.
cos(x2), sin(x2),
lnx
x
x
- i.nh ngh˜ıa 10.1.2. Tˆa.p ho..p mo.i nguyˆen h`am cu’a h`am f (x) trˆen
D
khoa’ng (a, b) du.o..c go.i l`a t´ıch phˆan bˆa´t di.nh cu’a h`am f (x) trˆen khoa’ng
y hiˆe.u l`a
(a, b) v`a du.o..c k´
f (x)dx.
Nˆe´u F (x) l`a mˆo.t trong c´ac nguyˆen h`am cu’a h`am f (x) trˆen khoa’ng
(a, b) th`ı theo di.nh l´
y 10.1.2
f (x)dx = F (x) + C,

C∈R

`an hiˆe’u l`a d˘a’ng th´
u.a
trong d´o C l`a h˘a`ng sˆo´ t`
uy y
´ v`a d˘a’ng th´
u.c cˆ
u.c gi˜
hai tˆa.p ho..p.
C´ac t´ınh chˆa´t co. ba’n cu’a t´ıch phˆan bˆa´t di.nh:
1) d

f (x)dx = f (x)dx.

2)

f (x)dx

3)

df (x) =

= f (x).
f (x)dx = f (x) + C.

T`
u. di.nh ngh˜ıa t´ıch phˆan bˆa´t di.nh r´
ut ra ba’ng c´ac t´ıch phˆan co.

ba’n (thu.`o.ng du.o..c go.i l`a t´ıch phˆan ba’ng) sau dˆay:


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

6

I.

0.dx = C.

II.

1dx = x + C.
xα+1
+ C, α = −1
α+1

III.

xαdx =

IV.

dx
= ln|x| + C, x = 0.
x

V.

axdx =

ax
+ C (0 < a = 1);
lna

VI.

sin xdx = − cos x + C.

VII.

cos xdx = sin x + C.

VIII.
IX.

X.

XI.

ex dx = ex + C.

dx
π
= tgx + C, x = + nπ, n ∈ Z.
2
cos x
2
dx

= −cotgx + C, x = nπ, n ∈ Z.
sin2 x

arc sin x + C,
dx
√
−1 < x < 1.
=
1 − x2 −arc cos x + C

arctgx + C,

dx
=
1 + x2 −arccotgx + C.

√
dx
= ln|x + x2 ± 1| + C
x2 ± 1
u. th`ı x < −1 ho˘a.c x > 1).
(trong tru.`o.ng ho..p dˆa´u tr`

XII.

XIII.

√

1 1+x

dx
+ C, |x| = 1.
= ln
2
1−x
2 1−x

C´ac quy t˘´ac t´ınh t´ıch phˆan bˆa´t di.nh:


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

1)

kf (x)dx = k

2)

[f (x) ± g(x)]dx =

3) Nˆe´u

7

f (x)dx, k = 0.
f (x)dx ±

g(x)dx.

f (x)dx = F (x) + C v`a u = ϕ(x) kha’ vi liˆen tu.c th`ı

f (u)du = F (u) + C.
´ V´I DU
CAC
.
V´ı du. 1. Ch´
u.ng minh r˘`ang h`am y = signx c´o nguyˆen h`am trˆen
khoa’ng bˆa´t k`
y khˆong ch´
u.a diˆe’m x = 0 v`a khˆong c´o nguyˆen h`am trˆen
mo.i khoa’ng ch´
u.a diˆe’m x = 0.
Gia’i. 1) Trˆen khoa’ng bˆa´t k`
y khˆong ch´
u.a diˆe’m x = 0 h`am y = signx
l`a h˘`ang sˆo´. Ch˘a’ng ha.n v´o.i mo.i khoa’ng (a, b), 0 < a < b ta c´o signx = 1
v`a do d´o mo.i nguyˆen h`am cu’a n´o trˆen (a, b) c´o da.ng
F (x) = x + C,

C ∈ R.

2) Ta x´et khoa’ng (a, b) m`a a < 0 < b. Trˆen khoa’ng (a, 0) mo.i
nguyˆen h`am cu’a signx c´o da.ng F (x) = −x + C1 c`on trˆen khoa’ng (0, b)
nguyˆen h`am c´o da.ng F (x) = x + C2. V´o.i mo.i c´ach cho.n h˘a`ng sˆo´ C1
v`a C2 ta thu du.o..c h`am [trˆen (a, b)] khˆong c´o da.o h`am ta.i diˆe’m x = 0.
Nˆe´u ta cho.n C = C1 = C2 th`ı thu du.o..c h`am liˆen tu.c y = |x| + C
u. d´o, theo di.nh ngh˜ıa 1 h`am
nhu.ng khˆong kha’ vi ta.i diˆe’m x = 0. T`
signx khˆong c´o nguyˆen h`am trˆen (a, b), a < 0 < b.
V´ı du. 2. T`ım nguyˆen h`am cu’a h`am f (x) = e|x| trˆen to`an tru.c sˆo´.

`en x > 0 mˆo.t
0 ta c´o e|x| = ex v`a do d´o trong miˆ
Gia’i. V´o.i x
trong c´ac nguyˆen h`am l`a ex . Khi x < 0 ta c´o e|x| = e−x v`a do vˆa.y
`en x < 0 mˆo.t trong c´ac nguyˆen h`am l`a −e−x + C v´o.i h˘a`ng
trong miˆ
sˆo´ C bˆa´t k`
y.
Theo di.nh ngh˜ıa, nguyˆen h`am cu’a h`am e|x| pha’i liˆen tu.c nˆen n´o


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

8
`eu kiˆe.n
pha’i tho’a m˜an diˆ

lim ex = lim (−e−x + C)

x→0+0

x→0−0

t´
u.c l`a 1 = −1 + C ⇒ C = 2.
Nhu. vˆa.y


ex
nˆe´u x > 0,



F (x) = 1
nˆe´u x = 0,



−e−x + 2 nˆe´u x < 0
l`a h`am liˆen tu.c trˆen to`an tru.c sˆo´. Ta ch´
u.ng minh r˘a`ng F (x) l`a nguyˆen
h`am cu’a h`am e|x| trˆen to`an tru.c sˆo´. Thˆa.t vˆa.y, v´o.i x > 0 ta c´o
`an pha’i
F (x) = ex = e|x|, v´o.i x < 0 th`ı F (x) = e−x = e|x|. Ta c`on cˆ
ch´
u.ng minh r˘`ang F (0) = e0 = 1. Ta c´o
F (x) − F (0)
ex − 1
= lim
= 1,
x→0+0
x→0+0
x
x
F (x) − F (0)
−e−x + 2 − 1
F− (0) = lim
= lim
= 1.
x→0−0
x→0−0

x
x
Nhu. vˆa.y F+ (0) = F− (0) = F (0) = 1 = e|x|. T`
u. d´o c´o thˆe’ viˆe´t:

ex + C,
x<0
e|x|dx = F (x) + C =
−e−x + 2 + C, x < 0.
F+ (0) = lim

`o thi. qua diˆe’m (−2, 2) dˆo´i v´o.i h`am
V´ı du. 3. T`ım nguyˆen h`am c´o dˆ
1
f (x) = , x ∈ (−∞, 0).
x
1
Gia’i. V`ı (ln|x|) = nˆen ln|x| l`a mˆo.t trong c´ac nguyˆen h`am cu’a
x
1
h`am f (x) = . Do vˆa.y, nguyˆen h`am cu’a f l`a h`am F (x) = ln|x| + C,
x
`eu kiˆe.n F (−2) = 2, t´
C ∈ R. H˘a`ng sˆo´ C du.o..c x´ac di.nh t`
u. diˆ
u.c l`a
ln2 + C = 2 ⇒ C = 2 − ln2. Nhu. vˆa.y
F (x) = ln|x| + 2 − ln2 = ln

x

+ 2.
2


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

9

V´ı du. 4. T´ınh c´ac t´ıch phˆan sau dˆay:
2x+1 − 5x−1
dx,
10x

1)

2)

2x + 3
dx.
3x + 2

Gia’i. 1) Ta c´o
2x
5x
1
dx =
2
−
x
x

10
5 · 10
5
1 x
1 x
1
=2
dx −
dx
5
5
2
1 x
1 x
1 2
=2 5
−
+C
1
1
5
ln
ln
5
2
2
1
=− x
+
+ C.

5 ln5 5 · 2x ln2

I=

x

2

−

1 1
5 2

x

dx

2)
5
2
3
+
x
+
2 dx = 2
3
6 dx
I=
2
2

3
x+
3 x+
3
3
2
5
2
= x + ln x + + C.
3
9
3
2 x+

V´ı du. 5. T´ınh c´ac t´ıch phˆan sau dˆay:
1)

tg2 xdx,

2)

1 + cos2 x
dx,
1 + cos 2x

3)

√
1 − sin 2xdx.

Gia’i. 1)
2

tg xdx =
=

1 − cos2 x
sin2 x
dx =
dx
cos2 x
cos2 x
dx
− dx = tgx − x + C.
cos2 x


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

10
2)
1 + cos2 x
dx =
1 + cos 2x

1 + cos2 x
1
dx =
2
2 cos x

2

dx
+
cos2 x

dx

1
= (tgx + x) + C.
2
3)
√
1 − sin 2xdx =

sin2 x − 2 sin x cos x + cos2 xdx
(sin x − cos x)2dx =

=

| sin x − cos x|dx

= (sin x + cos x)sign(cos x − sin x) + C.

` TA
ˆP
BAI
.
`ong nhˆa´t, h˜ay du.a c´ac t´ıch phˆan d˜a cho
B˘`ang c´ac ph´ep biˆe´n dˆo’i dˆ

`e t´ıch phˆan ba’ng v`a t´ınh c´ac t´ıch phˆan d´o1
vˆ
dx
.
x4 − 1

1.

(DS.

1
1 x−1
ln
− arctgx)
4 x+1
2

1
1 + 2x2
dx.
(DS.
)
arctgx
−
x2 (1 + x2 )
x
√
√
√
x2 + 1 + 1 − x2

√
dx.
(DS. arc sin x + ln|x + 1 + x2|)
1 − x4
√
√
√
√
x2 + 1 − 1 − x2
√
dx. (DS. ln|x + x2 − 1| − ln|x + x2 + 1|)
x4 − 1
√
1
x4 + x−4 + 2
dx.
(DS. ln|x| − 4 )
3
x
4x

2.
3.
4.
5.

23x − 1
dx.
ex − 1

6.

(DS.

e2x
+ ex + 1)
2

ay ch´
ung tˆ
oi bo’ qua khˆ
ong viˆe´t
Dˆe’ cho go.n, trong c´
ac “D´
ap sˆ
o´” cu’a chu.o.ng n`
`
c´
ac h˘
ang sˆ
o´ cˆ
o.ng C.
1


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

7.
8.

9.

11

3x

22x − 1
√
dx.
2x

2 22
x
+ 2− 2 )
(DS.
ln2 3

dx
.
x(2 + ln2 x)
√
3
ln2 x
dx.
x

1
lnx
(DS. √ arctg √ )
2

2
3 5/3
ln x)
5

(DS.

10.

ex + e2x
dx.
1 − ex

11.

ex dx
.
1 + ex

12.

x
sin2 dx.
2

(DS.

13.

cotg2 xdx.

(DS. −x − cotgx)

14.

√
π
.
1 + sin 2xdx, x ∈ 0,
2

15.

ecos x sin xdx.

16.

ex cos ex dx.

(DS. sin ex)

17.

1
dx.
1 + cos x

x
(DS. tg )
2

18.

dx
.
sin x + cos x

19.

1 + cos x
dx.
(x + sin x)3

20.

(DS. −ex − 2ln|ex − 1|)
(DS. ln(1 + ex))

sin 2x
2

(DS. − cos x + sin x)

(DS. −ecos x )

x π
1
(DS. √ ln tg
+
2

8
2

dx.

1 − 4 sin x
21.

sin x
1
x−
)
2
2

(DS. −

2
)
2(x + sin x)2

(DS. −

1
2

1 − 4 sin2 x)
√

sin x

2

2 − sin x

dx.

)

(DS. −ln| cos x +

1 + cos2 x|)


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

12

22.
23.
24.

sin x cos x
3 − sin4 x

sin2 x
1
(DS. arc sin √
)
2
3

dx.

1
arccotg3x
dx.
(DS. − arccotg2 3x)
2
1 + 9x
6
√
x + arctg2x
1
1
2
ln(1
+
4x
arctg3/22x)
dx.
(DS.
)
+
1 + 4x2
8
3

25.

arc sin x − arc cos x

√
dx.
1 − x2

26.

x + arc sin3 2x
√
dx.
1 − 4x2

27.

x + arc cos3/2 x
√
dx.
1 − x2

28.

x|x|dx.

29.

(2x − 3)|x − 2|dx.

30.

(DS.

(DS.

1
(arc sin2 x + arc cos2 x))
2

(DS. −

1√
1
1 − 4x2 + arc sin4 2x)
4
8

√
2
(DS. − 1 − x2 − arc cos5/2 x)
5

|x|3
)
3


7
2

− x3 + x2 − 6x + C, x < 2
3
2

(DS. F (x) =
)
7
2

 x3 − x2 + 6x + C,
x 2
3
2

1 − x2, |x| 1,
f (x)dx, f (x) =
1 − |x|, |x| > 1.

3

x − x + C
3
(DS. F (x) =

x − x|x| + 1 signx + C
2
6

10.1.2

ap dˆ
Phu.o.ng ph´
o’i biˆ
e´n

- i.nh l´
D
y. Gia’ su’.:

nˆe´u |x|

1
)

nˆe´u|x| > 1


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

13

a.p ho..p gi´
a
1) H`
am x = ϕ(t) x´
ac di.nh v`
a kha’ vi trˆen khoa’ng T v´
o.i tˆ
tri. l`
a khoa’ng X.
2) H`
am y = f (x) x´
ac di.nh v`
a c´

o nguyˆen h`
am F (x) trˆen khoa’ng X.
o h`
am F (ϕ(t)) l`
a nguyˆen h`
am cu’a h`
am f (ϕ(t))ϕ (t) trˆen
Khi d´
khoa’ng T .
y 10.1.1 suy r˘a`ng
T`
u. di.nh l´
f (ϕ(t))ϕ (t)dt = F (ϕ(t)) + C.

(10.1)

V`ı
F (ϕ(t)) + C = (F (x) + C)

x=ϕ(t)

=

f (x)dx

x=ϕ(t)

cho nˆen d˘a’ng th´
u.c (10.1) c´o thˆe’ viˆe´t du.´o.i da.ng
f (x)dx

x=ϕ(t)

=

f (ϕ(t))ϕ (t)dt.

(10.2)

u.c (10.2) du.o..c go.i l`a cˆong th´
u.c dˆo’i biˆe´n trong t´ıch phˆan
D˘a’ng th´
bˆa´t di.nh.
u. (10.2) thu
Nˆe´u h`am x = ϕ(t) c´o h`am ngu.o..c t = ϕ−1 (x) th`ı t`
du.o..c
f (x)dx =

f (ϕ(t))ϕ (t)dt

t=ϕ−1 (x)

.

(10.3)

`e ph´ep dˆo’i biˆe´n.
Ta nˆeu mˆo.t v`ai v´ı du. vˆ
√
u.a c˘an a2 − x2, a > 0

i) Nˆe´u biˆe’u th´
u.c du.´o.i dˆa´u t´ıch phˆan c´o ch´
π π
.
th`ı su’. du.ng ph´ep dˆo’i biˆe´n x = a sin t, t ∈ − ,
2 2 √
ii) Nˆe´u biˆe’u th´
u.c du.´o.i dˆa´u t´ıch phˆan c´o ch´
u.a c˘an x2 − a2, a > 0
π
a
, 0 < t < ho˘a.c x = acht.
th`ı d`
ung ph´ep dˆo’i biˆe´n x =
cos t
2
√
iii) Nˆe´u h`am du.´o.i dˆa´u t´ıch phˆan ch´
u.a c˘an th´
u.c a2 + x2, a > 0
π π
ho˘a.c x = asht.
th`ı c´o thˆe’ d˘a.t x = atgt, t ∈ − ,
2 2
.
.
iv) Nˆe´u h`am du ´o i dˆa´u t´ıch phˆan l`a f (x) = R(ex , e2x, . . . .enx ) th`ı
u.u ty’).
c´o thˆe’ d˘a.t t = ex (o’. dˆay R l`a h`am h˜

Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

14

´ V´I DU
CAC
.
V´ı du. 1. T´ınh

dx
.
cos x

Gia’i. Ta c´o
cos xdx
dx
(d˘a.t t = sin x, dt = cos xdx)
=
cos x
1 − sin2 x
x π
1 1+t
dt
ln
+
C
=
ln
tg

+
=
=
1 − t2
2 1−t
2
4
V´ı du. 2. T´ınh I =

+ C.

x3 dx
.
x8 − 2

Gia’i. ta c´o

√

I=

1
d(x4 )
4
=
x8 − 2

2 x4
d √
4

2
x4
−2 1 − √
2

2

x4
D˘a.t t = √ ta thu du.o..c
2
√
√
2
2 + x4
I=−
ln √
+ C.
8
2 − x4
x2 dx
·
(x2 + a2 )3
adt
Gia’i. D˘a.t x(t) = atgt ⇒ dx =
. Do d´o
cos2 t

V´ı du. 3. T´ınh I =

sin2 t

dt
a3tg2t · cos3 tdt
=
dt
=
− cos tdt
I=
a3 cos2 t
cos t
cos t
t π
+
− sin t + C.
= ln tg
2 4
x
V`ı t = arctg nˆen
a
1
x π
x
I = ln tg arctg +
− sin arctg
+C
2
a 4
a
√
x
= −√

+ ln|x + x2 + a2| + C.
x2 + a2


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

15

˜e d`ang thˆa´y r˘`ang
Thˆa.t vˆa.y, v`ı sin α = cos α · tgα nˆen dˆ
sin arctg

x
x
=√
·
2
a
x + a2

Tiˆe´p theo ta c´o
x π
1
arctg +
2
a 4
x π
1
cos arctg +
2

a 4
sin

x π
1 − cos arctg +
a 2
=
x π
sin arctg +
a 2
√
x + a2 + x2
=
a

x
1 + sin arctg
a
=
x
− cos arctg
a

`eu pha’i ch´
u.ng minh.
v`a t`
u. d´o suy ra diˆ
√
V´ı du. 4. T´ınh I =
a2 + x2 dx.

Gia’i. D˘a.t x = asht. Khi d´o
a2 (1 + sh2 t)achtdt = a2

I=
= a2
=

a2 1
ch2t + 1
dt =
sh2t + t + C
2
2 2

a2
(sht · cht + t) + C.
2
2

x2 t
x+
1 + 2 . e = sht + cht =
a

1 + sh t =
√
x + a2 + x2
t = ln
v`a do d´o
a

V`ı cht =

ch2 tdt

√
a2 + x2 dx =

√

a2 + x2
nˆen
a

√
x√ 2
a2
a + x2 + ln|x + a2 + x2| + C.
2
2

V´ı du. 5. T´ınh
1) I1 =

√

x2 + 1
dx,
x6 − 7x4 + x2

2) I2 =

3x + 4
√
dx.
−x2 + 6x − 8


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

16
Gia’i. 1) Ta c´o
1+
I1 =

1
x2

x2 − 7 +

d x−
1
x2

dx =
x−

1
x

1

x

√

=

2

−5

dt
t2 − 5

√
1
1
t2 − 5| + C = ln x − + x2 − 7 + 2 + C.
x
x
2) Ta viˆe´t biˆe’u th´
u.c du.´o.i dˆa´u t´ıch phˆan du.´o.i da.ng
1
3
−2x + 6
+ 13 · √
f (x) = − · √
2
−x2 + 6x − 8
−x2 + 6x − 8
v`a thu du.o..c

= ln|t +

I2 =

f (x)dx

1
3
(−x2 + 6x − 8)− 2 d(−x2 + 6x − 8) + 13
2
√
= −3 −x2 + 6x − 8 + 13 arc sin(x − 3) + C.

=−

d(x − 3)
1 − (x − 3)2

V´ı du. 6. T´ınh
dx
,
sin x

1)

2) I2 =

sin x cos3 x
dx.
1 + cos2 x

Gia’i
1) C´
ach I. Ta c´o
dx
=
sin x

sin x
dx =
sin2 x

1 1 − cos x
d(cos x)
=
ln
+ C.
cos2 x − 1
2 1 + cos x

C´
ach II.
dx
=
sin x

=

x
x

d
d
2
2
x =
x
x
x
sin cos
tg · cos2
2
2
2
2
x
d tg
x
2
x = ln tg 2 + C.
tg
2


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

17

2) Ta c´o
sin x cos x[(cos2 x + 1) − 1]
dx.

1 + cos2 x

I2 =

Ta d˘a.t t = 1 + cos2 x. T`
u. d´o dt = −2 cos x sin xdx. Do d´o
I2 = −

1
2

t−1
t
dt = − + ln|t| + C,
t
2

trong d´o t = 1 + cos2 x.
V´ı du. 7. T´ınh
1) I1 =

exdx
√
,
e2x + 5

2)

I2 =

ex + 1
dx.
ex − 1

Gia’i
1) D˘a.t ex = t. Ta c´o ex dx = dt v`a
√
√
dt
√
= ln|t + t2 + 5| + C = ln |ex + e2x + 5| + C.
t2 + 5

I1 =

dt
v`a thu du.o..c
2) Tu.o.ng tu.., d˘a.t ex = t, exdx = dt, dx =
t
I2 =

t + 1 dt
=
t−1 t

2dt
−
t−1

dt

= 2ln|t − 1| − ln|t| + C
t

= 2ln|ex − 1| − lnex + c
= ln(ex − 1)2 − x + C.

` TA
ˆP
BAI
.
T´ınh c´ac t´ıch phˆan:
1.

e2x
4
dx.
(DS.
(3ex − 4) 4 (ex + 1)3 )
x
21
e +1
˜
Chı’ dˆ
a n. D˘a.t ex + 1 = t4.
√
4


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

18

2.

dx
√
.
ex + 1

√
1 + ex − 1
(DS. ln √
)
1 + ex + 1

e2x
dx.
(DS. ex + ln|ex − 1|)
ex − 1
√
2
1 + lnx
4.
dx.
(DS.
(1 + lnx)3)
x
3
√
1 + lnx

dx.
5.
xlnx
√
√
(DS. 2 1 + lnx − ln|lnx| + 2ln| 1 + lnx − 1|)

3.

6.
7.
8.
9.

dx
x
x
.
(DS. −x − 2e− 2 + 2ln(1 + e 2 ))
x
+e
√
√
arctg x dx
√
.
(DS. (arctg x)2)
x 1+x
√
2

e3x + e2xdx.
(DS. (ex + 1)3/2 )
3
ex/2

e2x

2 +2x−1

(2x + 1)dx.

dx
.
ex − 1

10.

√

11.

e2xdx
√
.
e4x + 1

12.

2x dx
√

.
1 − 4x

13.

(DS.

1 2x2+2x−1
e
)
2

√
(DS. 2arctg ex − 1)
(DS.
(DS.

√
1
ln(e2x + e4x + 1))
2
arc sin 2x
)
ln2

√
√
dx
√
.

(DS. 2[ x + 1 − ln(1 + x + 1)])
1+ x+1
˜
Chı’ dˆ
a n. D˘a.t x + 1 = t2.

14.

x+1
√
dx.
x x−2

15.

√

dx
.
ax + b + m

√
√
(DS. 2 x − 2 + 2arctg
(DS.

x−2
)
2

√
2 √
ax + b − mln| ax + b + m| )
a


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

16.

dx
√
√
.
3
3
x( x − 1)

17.

dx
.
(1 − x2)3/2

√
√
(DS. 3 3 x + 3ln| 3 x − 1|)
(DS. tg(arc sin x))

˜

Chı’ dˆ
a n. D˘a.t x = sin t, t ∈
18.

dx
.
+ a2)3/2

−

π π
,
)
2 2

1
x
sin arctg )
2
a
a
π π
˜
.
Chı’ dˆ
a n. D˘a.t x = atgt, t ∈ − ,
2 2
(x2

19.

dx
.
(x2 − 1)3/2

21.

√
a2 + x2dx.

(DS.

1
1
, t = arc sin )
cos t
x
π
π
1
˜
Chı’ dˆ
a n. D˘a.t x =
, − < t < 0, 0 < t < .
sin t
2
2
√
√
x x a2 − x2

a2
20.
)
a2 − x2 dx.
(DS. arc sin +
2
a
2
˜
Chı’ dˆ
a n. D˘a.t x = a sin t.
(DS. −

(DS.

√
x√ 2
a2
a + x2 + ln|x + a2 + x2|)
2
2

˜
Chı’ dˆ
a n. D˘a.t x = asht.
22.
23.

x2
√

dx.
a2 + x2
x2

√

dx
.
x2 + a2

˜
Chı’ dˆ
a n. D˘a.t x =
24.

25.

√
1 √ 2
x a + x2 − a2ln(x + a2 + x2) )
2
√
x2 + a2
)
(DS. −
a2x

(DS.

1

ho˘a.c x = atgt, ho˘a.c x = asht.
t
x x√ 2
a2
(DS. arc sin −
a − x2 )
2
a a

x2dx
√
.
a2 − x2
˜
Chı’ dˆ
a n. D˘a.t x = a sin t.
dx
√
.
x x2 − a2

a
1
(DS. − arc sin )
a
x

19

Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

20

a
1
˜
ho˘a.c x = acht.
Chı’ dˆ
a n. D˘a.t x = , ho˘a.c x =
t
cos t
√
√
1 − x2
1 − x2
26.
− arc sin x)
dx.
(DS.
−
x2
x
27.

28.
29.
30.

dx

(a2

32.

.

dx
(x2

−

a2)3

(DS.

a2

x
√
)
x2 + a2

√
x2 − 9
)
(DS.
9x

dx
√

.
2
x x2 − 9
.

(DS. −

a2

√

x
)
x2 − a2

√
x2 a2 − x2dx.

(DS.

31.

+

x2 ) 3

x
x
a2 √
a4

− (a2 − x2)3/2 + x x2 − a2 + arc sin )
4
8
8
a

√
x
a+x
dx.
(DS. − a2 − x2 + arc sin )
a−x
a
˜
Chı’ dˆ
a n. D˘a.t x = a cos 2t.
x−a
dx.
x+a
√
√
√
(DS.
x2 − a2 − 2aln( x − a + x + a) nˆe´u x > a,
√
√
√
− x2 − a2 + 2aln( −x + a + −x − a) nˆe´u x < −a)
a
˜

.
Chı’ dˆ
a n. D˘a.t x =
cos 2t

33.

x − 1 dx
.
x + 1 x2

1
(DS. arc cos −
x

1
˜
Chı’ dˆ
a n. D˘a.t x = .
t
√
dx
√
34.
.
(DS. 2arc sin x)
x − x2

√
x2 − 1

)
x


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan

21

˜
Chı’ dˆ
a n. D˘a.t x = sin2 t.
√
√
√
1
+
x2 + 1
x2 + 1
dx.
(DS. x2 + 1 − ln
)
35.
x
x
36.

x3dx
√
.
2 − x2

(9 − x2)2
dx.
x6

37.
38.

(DS. −

x2dx
√
.
x2 − a2

(DS.

x2 √
4√
2 − x2 −
2 − x2 )
3
3

(DS. −

(9 − x2 )5
)
45x5

√

x√ 2
a2
x − a2 + ln|x + x2 − a2|)
2
2

(x + 1)dx
xex
)
.
(DS.
ln
x(1 + xex)
1 + xex
˜
`oi d˘a.t xex = t.
Chı’ dˆ
a n. Nhˆan tu’. sˆo´ v`a mˆa˜ u sˆo´ v´o.i ex rˆ
dx
ax
1
x
40.
+
.
(DS.
arctg
)
(x2 + a2)2
2a3

a x2 + a2
˜
Chı’ dˆ
a n. D˘a.t x = atgt.

39.

10.1.3

`an
ap t´ıch phˆ
an t`
u.ng phˆ
Phu.o.ng ph´

`an du..a trˆen di.nh l´
Phu.o.ng ph´ap t´ıch phˆan t`
y sau dˆay.
u.ng phˆ
- i.nh l´
D
y. Gia’ su’. trˆen khoa’ng D c´
ac h`
am u(x) v`
a v(x) kha’ vi v`
a h`
am
o nguyˆen h`
am. Khi d´
o h`

am u(x)v (x) c´
o nguyˆen h`
am trˆen
v(x)u (x) c´
D v`
a
u(x)v (x)dx = u(x)v(x) −

v(x)u (x)dx.

(10.4)

`an.
u.c t´ınh t´ıch phˆan t`
u.ng phˆ
Cˆong th´
u.c (10.4) du.o..c go.i l`a cˆong th´
V`ı u (x)dx = du v`a v (x)dx = dv nˆen (10.4) c´o thˆe’ viˆe´t du.´o.i da.ng
udv = uv −

vdu.

(10.4*)

`an l´o.n c´ac t´ıch phˆan t´ınh du.o..c b˘a`ng
Thu..c tˆe´ cho thˆa´y r˘a`ng phˆ
`an c´o thˆe’ phˆan th`anh ba nh´om sau dˆay.
ph´ep t´ıch phˆan t`
u.ng phˆ

Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

22

`om nh˜
u.a
Nh´
om I gˆ
u.ng t´ıch phˆan m`a h`am du.´o.i dˆa´u t´ıch phˆan c´o ch´
th`
u.a sˆo´ l`a mˆo.t trong c´ac h`am sau dˆay: lnx, arc sin x, arc cos x, arctgx,
(arctgx)2, (arc cos x)2, lnϕ(x), arc sin ϕ(x),...
u.c (10.4*) b˘a`ng c´ach
Dˆe’ t´ınh c´ac t´ıch phˆan n`ay ta ´ap du.ng cˆong th´
`an c`on la.i cu’a
d˘a.t u(x) b˘a`ng mˆo.t trong c´ac h`am d˜a chı’ ra c`on dv l`a phˆ
biˆe’u th´
u.c du.´o.i dˆa´u t´ıch phˆan.
`om nh˜
Nh´
om II gˆ
u.ng t´ıch phˆan m`a biˆe’u th´
u.c du.´o.i dˆa´u t´ıch phˆan
u.c, a,
c´o da.ng P (x)eax , P (x) cos bx, P (x) sin bx trong d´o P (x) l`a da th´
b l`a h˘a`ng sˆo´.
Dˆe’ t´ınh c´ac t´ıch phˆan n`ay ta ´ap du.ng (10.4*) b˘a`ng c´ach d˘a.t u(x) =
`an c`on la.i cu’a biˆe’u th´
P (x), dv l`a phˆ

u.c du.´o.i dˆa´u t´ıch phˆan. Sau mˆ˜o i
`an bˆa.c cu’a da th´
`an t´ıch phˆan t`
u.c s˜e gia’m mˆo.t do.n vi..
lˆ
u.ng phˆ
`om nh˜
Nh´
om III gˆ
u.ng t´ıch phˆan m`a h`am du.´o.i dˆa´u t´ıch phˆan c´o
`an t´ıch phˆan
da.ng: eax sin bx, eax cos bx, sin(lnx), cos(lnx),... Sau hai lˆ
`an ta la.i thu du.o..c t´ıch phˆan ban dˆ
`au v´o.i hˆe. sˆo´ n`ao d´o. D´o l`a
t`
u.ng phˆ
`an t´ınh.
phu.o.ng tr`ınh tuyˆe´n t´ınh v´o.i ˆa’n l`a t´ıch phˆan cˆ
.
.
.
u a nˆeu khˆong v´et hˆe´t mo.i t´ıch phˆan
Du o ng nhiˆen l`a ba nh´om v`
`an (xem v´ı du. 6).
u.ng phˆ
t´ınh du.o..c b˘`ang t´ıch phˆan t`
`an
u.ng phˆ
Nhˆ
a.n x´et. Nh`o. c´ac phu.o.ng ph´ap dˆo’i biˆe´n v`a t´ıch phˆan t`

ta ch´
u.ng minh du.o..c c´ac cˆong th´
u.c thu.`o.ng hay su’. du.ng sau dˆay:
1)
2)
3)
4)

x2

x
1
dx
= arctg + C, a = 0.
2
+a
a
a

a2

dx
a+x
1
+ C, a = 0.
= ln
2
−x
2a a − x

dx
x
√
= arc sin + C, a = 0.
2
2
a
a −x
√
dx
√
= ln|x + x2 ± a2| + C.
x2 ± a2


10.1. C´ac phu.o.ng ph´ap t´ınh t´ıch phˆan
´ V´I DU
CAC
.
√
√
V´ı du. 1. T´ınh t´ıch phˆan I =
xarctg xdx.
Gia’i. T´ıch phˆan d˜a cho thuˆo.c nh´om I. Ta d˘a.t
√
u(x) = arctg x,
√
dv = xdx.
Khi d´o du =

dx
2 3
1
· √ , v = x 2 . Do d´o
1+x 2 x
3

√
2 3
x
1
I = x 2 arctg x −
dx
3
3
1+x
√
1
2 3
1
1−
= x 2 arctg x −
dx
3
3
1+x
√
2 3
1
= x 2 arctg x − (x − ln|1 + x|) + C.

3
3
V´ı du. 2. T´ınh I = arc cos2 xdx.
Gia’i. Gia’ su’. u = arc cos2 x, dv = dx. Khi d´o
2arc cos x
du = − √
dx, v = x.
1 − x2
Theo (10.4*) ta c´o
xarc cos x
√
dx.
1 − x2
u.c thu du.o..c ta d˘a.t u =
Dˆe’ t´ınh t´ıch phˆan o’. vˆe´ pha’i d˘a’ng th´
xdx
. Khi d´o
arc cos x, dv = √
1 − x2
√
√
dx
du = − √
, v = − d( 1 − x2) = − 1 − x2 + C1
1 − x2
√
`an lˆa´y v = − 1 − x2:
v`a ta chı’ cˆ
√
xarc cos x

√
dx = − 1 − x2arc cos x − dx
21 − x2
√
= − 1 − x2arc cos x − x + C2 .
I = xarc cos2 x + 2

23


Chu.o.ng 10. T´ıch phˆan bˆa´t d .inh

24
Cuˆo´i c`
ung ta thu du.o..c

√
I = xarc cos2 x − 2 1 − x2arc cos x − 2x + C.
V´ı du. 3. T´ınh I =

x2 sin 3xdx.

Gia’i. T´ıch phˆan d˜a cho thuˆo.c nh´om II. Ta d˘a.t
u(x) = x2,
dv = sin 3xdx.
1
Khi d´o du = 2xdx, v = − cos 3x v`a
3
1
2

2
1
x cos 3xdx = − x2 cos 3x + I1.
I = − x2 cos 3x +
3
3
3
3
`an t´ınh I1. D˘a.t u = x, dv = cos 3xdx. Khi d´o du = 1dx,
Ta cˆ
1
u. d´o
v = sin 3x. T`
3
1
2 1
1
x sin 3x −
sin 3xdx
I = − x2 cos 3x +
3
3 3
3
2
2
1
cos 3x + C.
= − x2 cos 3x + x sin 3x +
3
9

27
`an t´ıch phˆan t`
Nhˆ
a.n x´et. Nˆe´u d˘a.t u = sin 3x, dv = x2dx th`ı lˆ
u.ng
`an th´
phˆ
u. nhˆa´t khˆong du.a dˆe´n t´ıch phˆan do.n gia’n ho.n.
V´ı du. 4. T´ınh I =

eax cos bx; a, b = 0.

Gia’i. Dˆay l`a t´ıch phˆan thuˆo.c nh´om III. Ta d˘a.t u = eax, dv =
1
cos bxdx. Khi d´o du = aeaxdx, v = sin bx v`a
b
1
a
1
a
I = eax sin bx −
eax sin bxdx = eax sin bx − I1 .
b
b
b
b
Dˆe’ t´ınh I1 ta d˘a.t u = eax, dv = sin bxdx. Khi d´o du = aeaxdx,
1
v = − cos bx v`a
b

1
a
eax cos bxdx.
I1 = − eax cos bx +
b
b