Giải nhanh bài toán nguyên hàm tích phân dành cho học sinh lớp 11 -12

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TUAN
ANH
m
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NGUYEN HAM
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Xufi't
ban nam 2013
La>i
noi dau
Viec
giai
mot bai toan noi chung la mot qua
trinh
tu duy cao do, dua tren
hilu
biet cua nguai
giai
toan. Viec
tinh
mot bai toan nguyen ham hay mot bai
toan
tich
phan cung vay. Co nguai tham chi khong
giai
dugc, c6 nguai
giai
dugc nhimg can qua
trinh
may mo rat lau, thu het
each

nay den
each
khac mai
giai
xong, trong khi c6 nguai lai tim dugc
each
giai
rat nhanh. Vay dau la bi
quyet de
giai
nhanh dugc mot bai toan nguyen ham, mot bai toan
tich
phan noi
rieng? Cach ren luyen de c6
each
giai
nhanh?
Cuon
sach
nay viet ra nhSm dem lai cho ban doe nhimg
each
hieu, nhijng
huang di, thu thuat de
tilp
can nhanh tai lai
giai
thoa dang cho mot bai toan
nguyen ham, mot bai toan
tich
phan. Cac cong thuc dua tai nguai doc khong

mang
tinh
ap dat ma theo huang de hieu, de nha de nguai doe c6 thien cam
han ve cac cong thuc do, phuc vu cho viec van dung
tinh
toan sau nay.
Cuon
sach
viet theo loi dien giang nen kho tranh
khoi
khiem khuyet, rat
mong nhan dugc nhung gop y thiet thuc
ciia
ban doe gan xa.
Xin
chan thanh cam an nhung gop y, chi dan cua quy thay:
-
TS. Nguyen Viet Dong, Truang Bo mon Giao due Toan hoc,
DHKHTN,
DHQG
TP. H6 Chi
Minh.
-
Thhy Nguyen Dinh Do, Pho
Hieu
truong Truang THPT Thanh Nhan -
TP.
Ho Chi
Minh.
-

ThSy Le Hoanh Sir, Giang vien
DHQG
TP.
HCM
- Truang DH
Kinh
Te Luat. i ' '
-
Thhy Nguyen Tat Thu, Giao vien Truang chuyen Luang Th6
Vinh
-
Bien
Hoa - Dong Nai.
Tran
Tuan
Anh
GIAI
NHANH BAI TOAN
TICH
PHAN TRONG
DE
THI TUYEN
SINH
DAI HOC NAM
2013
Cau
1;
Tinh tich
phan / = |—~
1"

•
(E>H kh6i A, Ai
- 2013)
Cdch
1: /=
——.lnxdx=
\nxdx
+
Cdch
gidi thong
thir&ng
2
w -I 1
In
xdx. Ta xet:
+
A
=
In
xdx.
Dat
u
= lnx=>
du
=
—dx
;
dv
= dx=>
v

=
x.
X
In
xdx
=
x\nx
2 '
1
-dx =
2\n2-\.
X
+
h =
-1
In
xdx
Dat
u
=
\nx^ du
=
—
dx; dv
=
X
-1
In
xdx
=

\X
J
f
1
\
,
1
rfX =>
V
= — .
X
1,
=
—
Inx
X
2
^
1
f
1
dx
X
=
—Inx
X
2
I
+
-

1
X
=
-ln2—.
1
2 2
Vay / = /,+/2=21n2-l + -ln2-i =
-(51n2-3).
V-1
.
In
xdx -
1-
1
Inxc/x
Dat
/•
=
In
X => X = e' va dx
=
e' dt.
Doi can:
x =
l=>/
= 0; x =
2=i>/
= ln2.
In
2/

, \n 2 In 2
1= ^\\—L-\e'dt=
\[e'-e")dt=
\td(e'+6-)
0 V
= t(e'+e-)
In
2
0
In 2
• '(e'+e-')dt
=
ln2.
V
2,
2-i
2
.l(51n2-3).
Vay
/
=
|(51n2-3).
Cac/i
^w/
nhanh
Cdch
3: cdc ban di y
quan
he
giita

— va \ : -^dx = d
X
X X
;
quan
he
giiia
X vd 1 la :
\dx
= dx.
Do
do, ta CO :
^x'-\^
dx
=
dx
= d
x
+ -
Vdy
ta
CO
the
giai
nhanh
bdi
todn tren
nhu
sau
:

2 2
1 =
x'-\
.
In
xdx
=
1-
1 ^
=
ln
X.
x
+ -
2 V O
-
x
+ -
1
rv
In
xdx
=
In
xd
*
1
X
x
+

-
= Inx.
\^2
x
+ -
2
1
+
1
^
dx
=
In
X.
r 0
2
( 1'
x
+
—
—
X
—
1
=
|(51n2-3).
I,^/ giai
that
nhanh ggn so
v&i

hai cdch tren !
Cau
2;
Tinh
tich
phan
/
=
^x^l-x'dx
(DH
kh6i
B - 2013
Cdch
giai thong thu&ng
Cdch
1: Do
dau
hieu
"
nen ta
chon
an
phu
x = V2
sin?.
Dat
X
=
y/l
sin

t =>
dx^-Jl
cos
tdt, te
n
n
Doi
can:
x = 0 / = 0;x =
1
=> / = —.
4
Taco:
/= JV2
sin
/
V2 - 2
sin"
/.72
cos /
Jr = 2 V2
Jsin
cos Vl -
sin^
/
J/
=
27^
sin
^ cos/.

cos/L//
=
2V2
sin/cos
tdt.
Xet
tich
phan
J = 272
|sin
/
cos^
tdt.
Dat
u =
cos/
^du =
-s'mtdt.
K
72
Doi
can
:
/
=
0 =o
w
= 1;/=
— =>
w

= —.
4
2
2 1 3
Taco:
J = -l4l
\u^du
= 2^
\u^du
= 2^.—
2
7^ =
272-1
Vay
/
=
272-1
Cdch
2:
Theo
kinh
nghiem thi thay
can
thuc
ta dat can
thuc
la
an
phu
!

Dat
/
= 72-x^ ^t^ =2-x^
=>tdt
= -xdx.
D6i
can:'
x =
0 =>
/
= 72;x =
1
=i>
/
=
1.
V2 2V2-1
Taco: I = -\t^dt= \t^^t= —
Cdch
gidi
nhanh
Cdch 3: Cdc ban de y quan he giua x vd x^ la:
xdx =
~^d{^x~^
=
-^d{2-x^y
Nen viec ta chon an phu t (0
cdch
2) la hodn todn tu nhien ! khong mang tinh dp dat cua kinh nghiem trong
suy nghi Id :

"thay
cd can
thiic
thi dat can
thitc
la an phu". Chung ta c6 the
gidi
nhanh nhie sau:
3
x42^dx
=
^\2-xjd(2-x')
=
^^^-p!-
1
_
2V2-I
0"
3
L&i
gidi
that
nhanh gon !
Cau
3:
Tinh
tich
phan
/ =
.(x

+
1)^
,
-<lx (DHkh6iD-2013)
x +1
Cdch
gidi
thong
thirong
Cdch 1: Ta c6 :
/
=
(x
+ 1)' + U V. V 2x
x'+\ x'+\
-dx = dx +
0
0
-dx = \
r
2x
-dx
Xet
tich
phan
J =
2x
x'+\
•dx.
Dat / = +1

=i>
J/ =
2xdx.
D6ican: x = 0 / = l;x =
1
/ = 2.
Tadugc
J= f—= ln
=
ln2.
Vay
/ = l + ln2.
Cdch 2: Ta c6 :
/=
;—'—dx= ; dx^ \dx+ \— dx^\
i
X +\ 1 J J r- -4-1 J
Xet
tich
phan
J =
2x
x^+1
x^+1
•dx.
0
0
2x_
x'+\
V

2x
x'+l
-dx
Dat x = tan/=>(ix =
—^—dt
= {\-^ian-t]dt, ti
cos"/ ^ '
Doican: x = 0 / = 0;x =
1
=^/ = —.
4
2
'2
Tadugc
/=
f^^(l
+ tan^/)^/ = 2 f^c//
tan / +
1
V
1
=
-2 :
of
(cos/) = -2 In
J cos/
1
sin/
<
cos/

cos/
4 =-2
In
0
4i'
Vay
/ =
l-2hi-^
= l + ln2.
V2
Cdch
gidi
nhanh
Cdch 3: Cdc ban de y quan he giua x vd x^ Id:
2xdx
= ci(x^ j = (x^ +1).
Nen viec ta chon dn phu t = x' -\-\(o cdch 1) Id hodn todn tu nhien ! Chung
ta
CO
the gidi nhanh nhu sau :
1 =
Ax + \)\+2x4-1
-flX
=
I
1
x^+1
,
dx = \dx + f ^"^ dx
0

0
=
X
+
x^+1
i/(x'+l)
=x
0
+
ln
x^+l
=
l + ln2.
LM
gidi
that
nhanh gon !
D6
CO
each
nhin
"tudng
minh" vh
each
giai
nhanh
Nguyen ham va
Tich
phan, mai ban doc tim hieu nhirng kien giai trong cuon
sach

nay !
ChLPcng
1.
NGUYEN
HAM
Bai
1.
NGUYEN
HAM
1. Dinh nghIa
Cho ham so f(x) xac
dinh
tren K (K la khoang ho^c doan
hoac
nua khoang
cua M). Ham s6 F(x) dugc goi la nguyen ham
ciia
ham s6 f(x) tren K nSu
F'(x)
=
f(x)
vai mgi x thugc K.
Mgi
ham s6 f(x)
lien
tuc tren K d^u c6 nguyen ham tren K.
Sau nay, yeu chu tim nguyen ham cua mot ham s6 dugc hieu la tim nguyen
ham tren tung khoang xac
dinh
cua no.

F(x)
la mot nguyen ham
ciia
ham f(x) thi F(x) + C (C la hang s6) la ho
nguyen ham cua ham f(x) hay
tich
phan hk
dinh
cua ham
f(x).
Ki
hieu :
fix)dx
=
F{x)
+
C
Vi
du 1
a)
J2xdx
=
x^+C vi
(x'+C)'
= 2x.
b)
cosxdx
=
smx
+

C vi (sinx +
C)'
= cosx.
*
Luu y: di hiiu nhanh nhung noi dung kien thuc trong cuon sdch nay, ban
doc
nen ren luyen thdnh thgo viec tinh dgo ham !
2.
Tinh
chat
thii-
nhat
f'{x)dx=fix)
+
C
Tinh
chat thu nhSt dugc suy true
tilp
tir
dinh
nghia nguyen ham. Trong thuc
hanh,
tinh
chk nay giup ta tim ra nguyen ham cua mot ham so don gian, cung
nhu
viec xac
dinh
lai nguyen ham tim ra c6 dung khong theo
each
nghi:

''muon
tim nguyen ham
ciia
ham so f(x),
chiing
ta tim ham so md dgo ham bgc nhat
cm no phdi chinh
la f(x)'\i
each
hieu do, chung ta c6 the thanh lap Bang
cong thuc nguyen ham co ban nhu sau :
(1)
Cong thirc 1 :
Qdx
=? Ta suy nghi : ham so nao c6 dao ham bac nhat
bang 0?
Hien
nhien do la hang so
!
Vay ta c6 cong thuc
thii
nhSt:
Qdx =
C
(2)
Cong thii-c 2 :
\dx=l
Ta suy nghi : ham so nao c6 dao ham bac nhat
bang 1? De dang nhan thay do la
X

vi x'
—
1. Vay ta c6 cong thuc thu
haii
\dx
=
=
x
+
C
(3)
Cong thii-c 3 : x"dx =? Ta suy
nghi:
ham s6 nao c6 dao ham bac nhat
bang jc"? Chung ta
lien
tuong ngay toi cong thuc dao ham {x")'
= nx"'^
hay
=
x"
. Ta thay n-\^a hay « = a
+1,
thu dugc cong thuc
«
+ l
=
X
hay
=

x" . Vay la ham so c6 dao ham bac nhat bang x". Suy ra
a
+
\
cong thuc thu ba :
a+\
x"dx=-—
+
C
ar
+
1
(a^-1).
(4) Cong thuc 4 : f—c/x =? Ta suy
nghi:
ham so nao c6 dao ham bac nhat
bang — Ta
lien
tuong toi cong thuc (inx) =— thi thu
duac
cong thuc
X
9 X
\-dx^\nx+C
J
V
.
Chiing
ta lay dau gia
tri

tuyet
doi vi
dieu
kien
ciia
ham
Logarit!
(5)
Cong thij-c 5 : a''dx=7 Ta suy nghi : ham so nao c6 dao ham bac nhk
bang a''? Tu cong thuc
tinh
dao ham quen thugc
(^a"^
=a''lna hay
=
a",
tiic
la ham so c6 dao ham bac nh^t bang a". Vay ta dl dang
In
a
vlna;
thu
dugc cong
thiic
a
a'dx
=
—
+
C

\na
(a>0,fl^l).
(6)
Cong thuc 6 : e''dx=? Ta suy
nghi:
ham so nao c6 dao ham bac nhSt
J
bang e"') De dang ta nhan thay do la ham e' vi (e'') =€' , suy ra cong
thiic
thu
sau : e^dx
=
6"
+
C . Cong thuc
thii
sau la truofng hgp rieng
ciia
cong
thiic
thii
nam khi thay "a" bang "e" !
(7) Cong thu-c 7 : jcosxdx =? Ta suy nghi : ham so nao c6 dao ham b$c
nhk bang
cosx?
Tir cong thuc quen thuoc
(sinx)
=cosx, ta c6 ngay cong
thuc thu bay la :
cosxi/x

=sinx
+
C
(8)
Cong thii'c 8 : sin xdx =? Ta suy
nghT:
ham so nao c6 dao ham bac nhat
bang sinx? Tu cong thuc quen thupc (cosx) =-sinx hay (-cosx) =sinx,
ta
CO
ham so ma dao ham bac nhat cua no bang "sinx" la "- cosx", suy ra cong
thuc thu tarn la :
|
sin
xdx = - cos x + C
(9)
Cong thuc 9 :
1
1
cos X
-dx=1
Ta suy nghT : ham so nao c6 dao ham bac
nhat bang
—r—
?
Truang hop nay khong de tim nguyen ham hon cac truoTig
cos X
hop tren ! chung ta dir doan ham so can tim c6 dang (chu y do mau thuc
cos
a;

"cos^x").
Ta c6:
cos
a;
A.
cosx +
sinx.
A
,
ro rang neu chon A = sinx thi
cos X
A.
COS
X
+ sinx.A cos^x + sin^o;
cos^ X
cos
X cos X

Vay ham so c6 dao ham bac
'
^ 1 '
Sill
X
nhat bang —-— la ham so hay tan x. Suy ra cong thuc thu
chin
:
cos X
cos a;
r

—Y~
— x + C
^
cos X
(10)
Congthiic 10
<-'
sin'
•dx = ? Ta suy nghi : ham so nao c6 dao ham
sm X
bac nhat bang
? Tuang tu cong thuc 9 !
Minh
du doan ham so can tim
CO
dang (chu y do mau thuc "sin^ x
").
Ta c6:
sinx
smx
la
sm^ X
sin^
X
—
sin^
X — cos^ X
sin^
X
^

. Vay ham so c6 dao ham bac nhat bang ^
sin^
X
sin^
X
ham so
-
cos X
sinx
hay (-
cotx).
Suy ra cong thuc thu muai:
sin^
X
dx = - cot X + C
Vay ta c6
Bang
nguyen ham ca ban sau :
Jodx
= C
f
adx = + C{cy >
0;
Q ^ 1)
In
a
J
dx = X + C
y
cos xdx

—
sin x + C
/
x"dx = - + C(a ^ -1)
J
sin xdx =
—
cos x + C
f
—
dx = \n X +C
X
f —^-— dx = tanx + C
^
cos X
J
e'dx = e'
+C
f
—— dx =
—
cot X + C
sin^
X
Hieu
va thuoc bang nguyen ham ca ban la dieu
kien
thiet yeu de chung ta
tinh
dugc nguyen ham cung nhu

tich
phan sau nay. Chinh vi vay, chung ta can
su dung thanh thao cac cong thuc trong bang nguyen ham ca ban.
3.
Tinh
chat thu- hai
J
kf{x)dx
= kj
f{x)dx
Trong
cong thuc nay, dieu ma chung ta can chu y la he s6 "k" (he so k c6
the "ra", "vao" qua dau nguyen ham!), tat nhien k phai la hang so, con bien so
khong
dua ra ngoai dSu nguyen ham dugc.
Vi
du 2. Ap dung
tinh
chSt thu hai va Bang nguyen ham ca ban, ta c6 :
a)
J 6xdx = GJ xdx (dp dung tinh chat thu hai)
=
6 — + C (dp dung bang nguyen ham ca ban)
2
=
3a;' + C .
,
. r cos X , If
1
cos xdx =

—
s
inx
+ C
•
3
c)
j
e^^'dx
= J e.e'dx = e
J
e'dx = e.e' + C.
d)
\mx\lx
= \Q\x'dx = \0.^—
+
C = 6x' +C.
-
+
1
3
Noi
chung, khi
tinh
nguyen ham cung nhu
tich
phan sau nay, chung ta c6
gang
bidn
d6i ham s6 dual dau nguyen ham hay dual dau

tich
phan xuat hien
nhung ham s6 c6 trong bang nguyen ham ca ban. Do vay, viec nam
dugc
Bang
nguyen ham co ban la di^u
kien
rSt quan trong de chung ta
tinh
dugc
nguyen
ham,
tich
phan.
4.
Tinh
chat
thir
ba
J" {J{x) ±
g{x))dx
=
J
f{x)dx
± J
g{x)dx
Chung ta c6 the hieu mot
each
dan gian cong
thiic

tren nhu sau: nguyen
ham
ciia
tong (hieu) cua hai ham so, bang tong (hieu) cac nguyen ham cua hai
ham so do.
Cong thuc
CO
the ma rgng nhu sau :
/
U^{x)±f^(x)± ±l{x))dx=^
f^{x)dx±
f
f^{x)dx± ±
J l{x)dx
Bay
gia
chiing
ta di xet cac vi du
minh
hoa :
Vi
du 3. Ap dung cac
tinh
chat
va Bang nguyen ham ca ban, ta c6 :
a)
= J {4x + 3cosa;)(ia; = J Axdx
+
J
3cosa;c?x-

(dp dung tinh chat
thu ba)
=
4 J" xdx +3
J
cos xdx (dp dung tinh chdi thu hai)
=
4 h
3
sin
a;
+ C (dp dung bang nguyen ham ca ban)
•
2
=
2a;^ + 3 sin X + C
b)
/ = r
(5e"
^)dx = r
Se'dx-
f dx (dp dung tinh chdt
'
cos' x *^
thif
ba)
cos' x
=
5
r

e^'dx
- 7 r
—-—
dx (dp dung tinh chdt thu hai)
.
"J
^ cos' X
=
56^^-7 tan x + C . (dp dung bang nguyen ham ca ban)
^
' . f 1 .
'^-^-Vx+
[x'dx
c)
I, =: j 3- +^ c/x= J3-Vx+ \-jr^dx = \l>\y
y <lx J \x
I
=
9{ydx+
{x'dx = — + ^ + C = -—
+
3x'+C.
J
J ln3 1 ln3
3
3x'
- 2x + 4
dx =
X
I

Sx' 2x 4
+
-
XXX
dx
J
Sxdx - J 2dx + j
:
3Jxdx-2Jdx +
AJ-dx
^ — -2x + A\nx + C.
X
2
Trong
thuc hanh, ta
trinh
bay nhanh nhu sau :
a) I,
=
Ji4x
/(5e
c)
I,
\
J 33;'
3x'
-^)dx
=5e^ - 7 tan x + C.
cos X
-dx

X
3\3'+x'
-2\ i
cix
= —
+
3x'+C.
in
3
3x
- 2 + -
X
dx
23;
+ 4 hi X + C.
Tiiy
theo kha nang cua nguai lam todn met ta c6 the lucre bo di nhirng buac
gidi khong can thiet.
Vi
du 4.
Tinh
:
^ X
c)
= j
T.ydx;
d)
h =
X
dx

-dx.
42
Gidi
Ta
bien doi ham so dual ddu nguyen ham ve dang ham cd chira cdc ham
trong
bdng nguyen ham co bdn de
tinh.
a) Ta CO :
-I
X
+ 1)^
dx
X
-I
X
dx
X
1
+
2 1
2 1
1
+ —+ -
X
^
dx = X- + 4^2!^ +
In
X + C.
dx

b)
Ta CO : 7^ = J
X
+ - e^x^
•dx
x
x e'x^
3 3 ,3
JC JC
dx
-I
1 1
x X
dx = —- + In
X
X
~e'' +C.
c)
Taco
': 73 =
JT.2,'dx
=
J(2.3)'dx
=
jG'dx
= — + C.
In
6
re
d)

Taco
: =
\—;<ix
=
4'
42
V /
-dx =
ce
-dx-^ dx =
e
v2y
V
^ /
In
'e^
v2/
+ C.
Vi du 5. Cho ham s6 f{x) = xe' va F{x) = {ax + b)e''. Vai gia
tri
nao cua a va
b
thi la mot nguyen ham ciia f(x) 9
Gidi
Tap xac
djnh
cua F(x) va /(x) la R. Ham s6 F(x) la mpt nguyen ham
cua f(x) thi
/^'(^)
= fix) vai Vx e R .

Ta CO : F'(x) = ae" + (ax + b)e''(ax + a + b)e" nen
F'(x)=-f(x)
vai
Vx e R thi (ax + a + b)e' = ' vai Vx e R <=> ax + a + 6 = x, Vx e R
a-l = 0 |'a = l
<^(o-l)x
+ a +
Z)-0,VxeR
<=><^ ,
Ci><
. Vay vai a = l va
a + b-0 b = -l
b l thi F(x) la mot nguyen ham cua f(x).
Vi
du 6, Chung
minh
rang F(x) - sin
xe""
la mot nguyen ham cua ham so
/(x)
= (sin X + cos xy .
Gidi
Tap xac
djnh
cua F(x) va /(x) la IR .
Taco: F'(x) = (sinx)'^''+ sinx(e'')'
= cos
xe""
+ sin xe' = (cos x + sin x)e'' = /(x).
Vay F(x) = sin xe' la mot nguyen ham cua ham so /(x) = (sin x + cos

x)e''.
(dpcm)
BAITAP
1.
Tinh
:
25a.''
+122'+ 1991
X
dx;
X
dx; b) 7, = J
15a;
+ 10V^ + 1983
da;.
2.
Tinh
:
a)=
J'^Ssina;
—
4cosa;jda;;
b) 7^ = J"
tan
a;
a;'
-
2a;
+ -
X

da;.
cos X sm X
dx
. 'J SI
sin
2x
-dx',
3.
Tinh
:
a) / = / • e'dx,
c)
=
JZ'e^dx;
4.
Tinh
:
X
—
xe
X
dx
dx.
r(VJ+4^
+ b^)dx; b) / = ++-1=
^)dx
5. Cho ham s6 /(x) = (x' + x)e" va
F(JC)
= (ax' +bx
+

c)e'. Vai gia
tri
nao
ciia
a, b va c thi F{x) la mot nguyen ham cua /{x) ?
Bai
2.
BANG NGUYEN HAM
MQ
RONG
Sail
day chung ta se ma
rong
cac cong thuc nguyen ham ca ban de dugc
Bang
nguyen ham ma
rong.
Bang
nguyen ham ma
rong
la cong cu giup chung
ta
tinh nhanh nguyen ham va
tich
phdn.
Truac
tien ta xet dinh li sau :
1. Dinh li
Nh J f(u)du =F{u) + C vau = u(x) la ham s6 c6 dgo ham
lien

tuc thi:
J
f{u(x)).u\x)dx
=F{u{x))
+ C
2. Cong thirc nguyen ham
mo" rOng
Ap
dung
dinh
li
tren trong truong hop u
=
ax
+
b{a^0),t^c6
:
jfiax
+
b)dx=
jf{ax
+
b).{ax
+
by dx
a
f(ax
+
b).{ax
+

bydx
=-F{ax
+
b)
+
C
Tom lai,
ta c6 cong thuc dk mo rong bang nguyen ham ca ban:
'/{ax
+
b)dx
=
-F(ax
+
b)
+
C
a
Cong
thuc nguyen ham ca ban va cong thuc nguyen ham ma rpng
dugc
cho
tuang ung duai bang sau :
Cong
t/iii'c
nguyen
ham
cff
ban
Cong

thuc
nguyen
ham
m&
rpng
Jodx
= C
Jdx
= x + C
f
x"dx = ^^ + C{a * -1)
J
a + 1
f{ax+bfdc='^.^'^^^^'^\c{a^
1)
a a+l
f
-dx = hi
1
X
1
+C
J
X
f
—-—
dx =
—
An
\ + b \

^
ax +
b
a
J
e'dx = +C
f
e'"'-'dx
= -e'^^'' +C
'J
a
f
adx = — +
C{a>Q]a^
1)
In
a
ra'^^'dx = ±.^
+ C{a
>
0;a
^
1)
^
a \na
j
cos xdx = sin x
-\-
C
/

cos (ax + b)dx =
—
sin
(ax +
b)
+ C
J
sin xdx —
—
cos x + C
f
sin(ax +
b)dx
=
— —
cos(ax
+b) + C
a
f
—\
dx — tanx + C
^
cos X
r
I 1
1
— dx =
—
ta n{ax
+

b)
+ C
^
cos (ax +
b)
a
f
—\ dx
=
—
cot x + C
^
sin X
r
1 1
/
— dx = cot(ax +
6)-1-C
sin (ox+ 6)
a
Trong
thuc hanh
tinh
nguyen ham cung nhu
tich
phan sau nay, a nhieu
truang hop viec ap dung bang nguyen ham ma rong cho ta lai
giai
bai toan
nhanh va "

sang
" han ! Chang han vai bai toan sau :
Tinh
nguyen ham : I
—
J (2x + l^dx. Neu khong ap dung cong thuc
nguyen ham ma rong thi ta khai
trien
bieu thuc {2x +
1)^,
sau do mai ap dung
cong thuc nguyen ham ca ban de
tinh
:
I
= J{2x +
lydx
= J(2x + l)(2x + Ifdx
= J
{2x
+
l)i8x''+12x^+6x
+ l)dx
= J
(16a;'
+ 24a;' + 12a;' +2x + 8x'' + 12a;' + 6x + l)dx
= J (16x' + 32a;' + 24a;' + 8a; + l)dx
16x'
32x* 24x' 8x'
+

Sa;" + 8a;' + 4a;^ + a; + C.
16x-'
5
Bay gia chung ta ap dung cong thuc nguyen ham ma rong (cong
thuc r{ax +
bydx
=
^"•'^
^ + C(a ^ -1)) d^
tinh/,
ta c6 :
^
o a +1
«^
^ ^ 2 4 +
1
10
(C/zw
>•
rang,
each
nay va
each
tren
deu cho kit qua
dung,
no chi sai
khdc
nhau
mot

hang
so xde
dinh!)
Neu
bai toan tren ta thay s6 mu 4 bang so mu 2013 chang han thi lai giai
nhu
each
dSu tien se phuc tap nhu the
nao?
Con neu chung ta ap dung cong
thuc nguyen ham ma rong ta c6
ngay
lai giai ngan gon cho bai toan do la
/
= / (2a; + If^^dx = ^ '— + C. Ke ca chung ta dung phuang phap
J
4028
d6i
biln
s6 (se hoc a bai sau) thi cung c6
lofi
giai khong gon bang
each
nay !
R6 rang cong thuc nguyen ham ma rong to ra uu diem han cong thuc nguyen
ham ca ban ! Cac cong thuc nguyen ham ma rong, neu chung ta cho he so a =
1;
i
=
0 thi ta thu

dugc
cong thuc nguyen ham ca ban.
Vi du 1.
Tinh
:
a) = J (2a; + s)'' dx; b) = J'(3 - 2a;)' dx
(3i-l)
^(33;-2)
dx
•
Giai
Ap
dung cac cong thuc
trong
Bang nguyen ham ma rong ta c6 :
/
\13 / vl3
c, xi2 1 2a;+ 3 2a; + 3
a) / = r 2a; + 3 dx = ^.^ = ^ -L-
'
^ J \ 2 13 26
r, v3 1 (3-2a;f
-(3-2a;)'
b) / = / 3-2a; dx = ^ L + =
__V
L
+
C.
r
I ci \-7 1

(3a;-l]
c) / = / -(^a; = / (3a;-l dx = -> '—^ C
'
^
(3: lf
^ ^ ^ 3-6
-1
18(3a;-l)
-7
+ C-
d) /, = {•
,
dx=
J^A:
= 2
{3x-2)idx
3' -5
+
1
*
Nil
an
xet:
-
Chung
ta c6 the
trinh
bay
nhanh
nhu sau :

2a;+
3)
a)/,=/
12
23;+
3 dx =
vl3
26
3
—
2xI dx =
-(3-2.)'
8
dx
1
-5
2 .,,.2_(3x-2)3
18(3a;-l)'
-1
3
-
D6i vai Cdu c) vd cdu d), neu sir
dung
cong
thitc
#x-2)^
+
C.
•dx = —
-1

a (n-l){ax + b)
—
+ C{n^l;a^O)
se cho
ta lai gidi nhanh han nua (vi gidm duac mot
buac
biin doi) !
-1
(3x-l)
18(3a;-l)
+
C.
d)L
=
rdx =
-
-1
•
+ C.
-
Neu khong dp dung cong thuc nguyen ham ma rong, chung ta gidi bdi
todn
bang
phuang phdp doi bien so md chung ta se xet trong bdi hoc sau. O
day chung ta dan cu cdu a) duac gidi bdng phuang phdp doi bien so de so
sdnh hai cdch gidi:
/
= 2a; + 3 dx
12
1

Dat
M = 2a: + 3 =>
(iti
= 2dx rfx = - du
2
Ta
CO
5:/ = \v}^-du = — + — + C
1 J r, cy -
2
13 26
Thay
u = 2x + 3, ta duac: /, = ^ + C.
•
^ 26
Rd
rang cdch nay to ra khd phuc tap so vai mot bdi todn dan gidn nhu vdy !
Ap
dung cong thuc nguyen ham ma rong ta c6 lai gidi gon vd nhanh han:
\13
12
2a; + 3 dx =
2x + 3
26
+
C
Do
vdy, viec nha vd van dung tot cong thuc nguyen ham ma rong Id can
thiet de chung ta tinh nhanh duac nguyen hdm vd tich phdn sau nay.
Vi

du 2.
Tinh
:
a)
/, =
f—^dx;
1
—X
c)
h= e ^ dx;
b)/,=/3
d)
7^
=.
J e-'dx
dx
Gidi
Ap
dung
cac cong
thuc
trong
Bang
nguyen
ham ma
rong
ta c6 :
1
a) 7^= f-^dx = —In
2-5a; -5

2
—
5a;
+
C =
—In
5
2-5a'
+
C.
b)
7 = f S'^'^dx = 3 r3'^ia; = S —+ C =+ C•
'J J 5 ln3 5 ln3
c)
l3= \e ^''dx = ^e ^''+C = -3e +C.
3
d)
7 =
fe-''dx
=
—e-'+C
=
-e-^+C.
-
Chung ta c6 the trinh bdy nhanh nhu sau :
2-5x
+
C a) 7 = / —-— dx =
—^
In

'
J 2-5a; 5
b)
7 =
r3'^'-'dx
- f 3.3'^rfa; = —- + C
'J J 5 In 3
_1
_1
c)/3= \e 3^c/x = -3e ^\c.
Vi
du 3.
Tinh
:
a)7j = J sin(3-4a;ya;; b) 7^
=
J 5cos{—^)dx
:)
I = f-L-
cos^(3;r)
•dx'
sin
(2z-l)
•dx-
Gidi
Ap
dung
cac cong
thuc
trong

Bang
nguyen
ham ma
rong
ta c6 :
a) /j =
Jsin(3
-
Ax)dx
= ^.(-cos(3 - 4a;)) + C
cos(3 - 4a;) ^
4
b) h = JScos
\ /
dx =
5.—.sin
+
C 15sin
+
C
\ /
c)
I = f dx = f ^ dx =
-temSx
+ C-
cos\3x)
3
^)
I. =
cos'(3.x)

1
1
3
-dx =
3.( cot(2a;
- 1)) + C
sin^(22;-l) 2
=
-|cot(2a;-l) + C-
-
Chung ta c6 thi trinh bay
nhanh
nhu sau :
a) / = fsin{3-4x)dx=:^^^^^^-^C.
J
4
b) = J 5cos{^)dx = -15sin(^) + C
c) / = f
dx^-tanSx
+ C-
'
J cos\3x) 3
^
ail
^
da; =-|cot(2x-l) + C.
sin'(2x-l)
BAITAP
1. Tinh :
a) =

J
(4x + 2)' dx;
b) ^. = /-^^-;
^
(3-2x)
^(41 + 5)
2. Tinh:
c)/2= |e 3 Jx;
b) = J
4 J
3. Tinh :
r(sin(3 ) +
cos53;)t/x;
b) / f ^
2 ^ cos (4
—
x)
1
.a:
—
h sm
—
sin 3a; 2
;
d)
^4
=/
2cos(3 )rfa;.
Bai
3.

PHl/QNG PHAP DOI
BIEN
SO
Bai
nay chiing ta se xet hai truong hop Ichi tinh
nguyen
ham /{x)dx
bang
phuong
phap
doi bien so :
-
Truong hop 1 : Dat u Id mot ham so cua x.
—
Truong hop 2 : Dat x la mot ham so cua u.
A.
Phep
dat u la mot ham so cua x : u = u(x)
Gia su can tinh / = /{x)dx, ta
thuc
hien nhu sau :
Buoc 1 : Chon an phu thich hop u
=
u(x).
Buoc 2
:
Xdc dinh viphdn du
=
du(x)
hoac

du^ - du^(x)
Buoc 3 : Bieuthi f(x)dx
theo
u va du. Gidsurang
f
(x)dx
=
g{u)du.
Buoc 4 : Tinh I
=
g(u)du. Sau do
thay
u = w(x) de
dime
ket qua can tim.
Chu V
:
chon
an phu u = w(x) sao cho
viec
tinh / = g(u)du phai de hon la
tinh
/= jf(x)dx !
Khi
nhin vao mot bai giai cho bai
toan
tinh
nguyen
ham hay tich
phan

bang
phuong
phap
dat an phu (hay
phuong
phap
doi bien so), ban doc
thuong
c6 cau
hoi
: tai sao lai
chon
dat an phu nhu
vay?
Lam sao
chon
an phu thich
hop?
Nhirng Icien
thuc
duai day se giiip cae ban dinh
huong
dugc
phep
dat an phu
cho minh mot
each
nhanh
chong
ma Ichong phai may mo lam giam toe do tinh

nguyen
ham, tich
phan
cua cac ban.
Truoc
tien cac ban c^n luu y hai ket qua ma chiing ta
thuong
dung
sau day :
(1) df{x) = f\x)dx .
(2) Niu J f(u)du =F{u) + C va u = u(x) la ham sS c6 dqo ham lien tuc
thl:
J
f{u{x)).u\x)du
=
J
i\u{x))du{x)
=F{u{x))
+ C
Vi
du
a)
J cos(2x^ + 3a; +
l)d{2x^
+ 3a; + 1) = sin(2a;^ + 3a,' + 1) + C.
(ta
hieu trong suy nghi "
2x^
+ 3x + 1 " lau)
b)

f ^- d(x' + 1) =ln a;' +
1
+ C = ln(a-' + 1) + C v/ + 1 > 0 (ta
^
(a;' +1)
hieu trong suy nghi " -\-\" la
u
)
Sau day chiing ta tim hi6u cac moi quan he quan trpng giup chung ta tim
nhanh
phep
dat in phu va dinh huong nhanh
each
giai
cho bai toan nguyen
ham,
tich
phan
bang
phuang
phap
doi bien so.
l.Quan
hegiua x" va
x"*\n^-\)
Ta
CO
: dx'"'
=
(«

+
\)x"dx o x"dx
=
—dx"*' '
n
+
\
+
\)
d{ax"^'+b),
trong
do a^O con b tuy y tren R . Vay ta c6 quan he
giira
x" va
x"^\n^-1)
nhu
(ta
hieu cong thicc tren mot each don gidn
sau :
x"dx =
-
1
a{n +
\)
d{ax"''+b)
nhu sau : dua x" vdo trong vi phan thl thdnh
{ax"*^
-^b), voi a ^ Q vd h tuy
ytren
Vidu

l.Tinh
:
a)
Jx^{2x^
+lfdx\ =
Jx4^+
Idx
Giai
a)
Phan
tich
hai
todn:
Theo I6i giai thong thuang, cac ban se khai trien
bieu thuc (2x + 1) , sau do nhdn vai X de dua ve nguyen ham de tinh han.
The
nhung viec khai trien bieu thicc (2x^ + if la khong dan gidn? Do vdy,
each
nay da to ra khong hieu qua ! Niu giai bai todn nay bang phuang phap
doi
biin so, ta chon an phu la u-lx^
+
\. Tgi sao Igi chon duac an phu nhu
vay?
Bay gia cac ban de y quan he
giica
x^ va x^ nhu sau :
x^dx =
-d(2x^ +1), nen ta c6
x'{2x^

+ ifdx =
-(2a;'
+
lfd{2x^
+ 1).
6 6
Do
do, viec chung ta chon an phu la u-2x^
+
\ hodn todn ty nhien, khong
mang tinh dp dat.
L&i
giai
cua bai
todn
J
x'(2x'
+ Ifdx =
J-(2x'
+
lfd{2x'
+ !)•
Dat u = 2x^ +1 du = d(2x^ +1).
Taco: T = - / uMu = + C = + C-
'6-^
6 10 60
Thay u = 2x^+1 ta
duac:
/ +
'

60
b)
Phan
tich
bai
todn:
Cac ban de y quan he
giica
x vd x :
xdx
=
- d(x^ +1) nen ta c6 x^x^ +\dx = -
Vx^
+1 d(x^ +1). Do vdy, ta c6 thi
2 2
chon an phu Id
w
= +1 hodc u
= Vx" +1.
Trong
truang hap nay ta nen chon
u = de bieu thuc dual ddu nguyen ham khong con can thuc
L&i
giai
cHa
bai
todn
fxyjx'+ldx
=
J^Vx'+l

d(x' + l).
Dat u =
Vx'+l
^u'
=x'+l:^du'
=:d(x'+l).
1 1 ^
Taco: = Jiu du' = J-u 2udu =
JuMu
=•— + 0-
/"I—T
t A (Vx^ + 1 )
Thay u ^ Vx +1 ta dugc: i =
'
3
Cach
khac :
'
Dat u = x^+l=:>du =
d(x'+l).
3
Ta c6:
I2
= j-Judu
=
—
ju^du =
+
C
= ^^^^^

+
C
2 2 2 3
2
(Vx^
+1 f
Thay u x +1 ta duac: I = 1- C •
' 3
* Nhgn xet: Neu da thanh thgo trong viec sic dung phuang phdp nay, cdc
ban
CO
thi trinh bay loi gidi nhanh hon nhu sau :
a)
= J x'(2x' + l)Mx = J
-{2x'
+ l)M(2x' + 1)
= + C • (ta hieu trong suy nghi "2x^ + 1" la "u")
60
b)l2= \x^x^+ldx =
-4x^+\d{x^
+1)
=
- f(x^ +1)2d{x^ +1)
. +
C
• (ta hieu trong suy nghi " x +1" la "u")
2.
Quan giua va —
Ta CO
'1'

nen quan he can xet giila — va ^ la:
X X X
(ta hieu cong thuc tren mot each don gidn nhu sau : dua \ trong vi
phdn thi thanh — +
b,
voi a ^ 0 vdbtuyy tren R)
X
Vi
du 2. Tinh :
l4
re ^
a)
Ii = \—r-dx;
X
r I 1 1
b) / = I — sin — cos — dx.
^ X X X
Gidi
a) Phdn tick bdi todn : Niu chua dugc biet din quan he giua \ — thi
X X
that khong de de chung ta tim ra ngay phep dgt an phu! Cdc ban de y quan he
.,1,11,
-1 ,
giua — va — ; —
dx
= — a
X X 3
1+-
X
nen ta co

, e
-1 3^
— d\-—e ^d
1
+ -
Do do, ta
CO
thi chon an phu la
w
=
1
+ —.
X
L&i gidi cua bdi todn
1.1
-dx =
f-l
1+- r 3^
e ^d
1
+ -
Dat u = ! + -=>du = d(l +
-).
X X
Ta CO
: I = —
reMu-
—e"
+C.
3 -1

Thay u =
1
+ — ta duac: Ii = —e ^ + C.
X
• 3
b) Phdn tick bdi todn : Cdc ban de y quan he giua va —
X X
1
dx = —d
ta
CO
the chon an phu Id u
=
— .
X
.,.1.1 1 , .1 1 ,
nen ta co — sm — cos — dx — - sin — cos —
d
XXX XX
1
. Do do,
L&i gidi cua bdi todn
1 r • 2 J
1
= / sin —d
—
X
2^
X
X

Dat u = i =^ du = d
X
Taco:
i _i f sin(2u) du =-i( cos2u) + C cos2u + C-
2 9.J ^ 2 2 4
Thay u = — ta duac: I = — cos
X 2 4
+ C-
* Nhan xet: Niu dd thanh thao trong viec sic dung phuang phdp nay, cdc
ban
CO
thi trinh bay lai gidi nhanh han nhu sau :
3
1+-
a)Ii
=
-dx =
\ /
1+
e ^of
-1 1+-
3
b) 1 = / — sm — cos — ax = — i sm
—cos
—a — = f sin —a
"J
X X X XXX 2^ X
=
—
cos

4
1
3. Quan he giua — va line
X
Ta CO (In x) =
—
nen quan he can xet giiia — va In x la :
— dx = — d{a \nx + b)
X
a
(ta hieu
cong
thuc tren mot each dan gidn nhu sau : dua — vdo trong vi
X
phdn
thi thanh {a In x
-\-h),
vai a ^Ovdb tiiyy tren R)
Vi
du 3. Tinh :
2
In
X + 3
-dx;
X
b)I,=/
21n'x + 5ln'x
xlnx
dx.
Gidi:

a)
Phan tich bai todn: Cdc ban di y quan he giita — va In x ;
X
^dx =
^d(21na,
+ 3) nen ta c6 ill!lfl2La!x =
l(21njc
+
3)V(21nx
+
3).
Do
vay, ta chon an phu Id u 2lnx + 3.
Ldi
gidi cda bai todn
n (2
In
X + pi,
>
9
Ij
= J ^dx =
J^(21nx
+ 3)
d(21nx
+ 3).
Dat u = 2
In
X + 3 => du = d(2
In

X + 3).
Ta CO
6:1
=ifuMu=1.^:^+0=^+0.
1 2 10 20
Thay u = 21nx + 3 tadugc: / ^
(2Inx
+ 3)^" ^ ^
' 20
b)
Phdn tich bai todn: Cdc ban dSy quan he giua — va \nx : —dx = d{\n x)
x X
2ln^x + 51n^x , 21n^ x + 51n^ x \ , , ^ ,
yi^n
ta
CO
dx = d[\n x). Do vay, ta chon an phu
xlnx
/fl
w
=
In
X.
Inx
LM
gidi cda bai todn
21n'x + 5ln'x
r21n-'x
In
X

dx
2
In^
X + 5
In^
x
Dat u = Inx => du = d(lnx).
•2u'
+ 5u'
Taco:
= j:
du
Inx
2u'
5u'
+ —
d(lnx).
u
u
du
= J
(2u^
+ 5u) du
3 2
Thay u = Inx ta
duac
: i ='^i^RJ^ +
^A\}12^
+ c
•

' 3 2
* Nhan xet: Niu da thanh thao trong viec su dung phuang phdp nay, cdc
ban
CO
the trinh bay lai gidi nhanh han nhu sau:
2
In
X + 3 p \ K9
i
^dx=
/•-(21nx
+ 3
d(21nx
+ 3)
a) 1,=/
_(21nx
+ 3r ^ ^
20
., r 2
In^
x + 5
In^
X , r 2
In^
x + 5
In^
x ,,
b) I / dx = / ; d(lnx)
xlnx
^

Inx
"-/, ^2 ^, 1,., X 2(lrix)^ 5(lnx)^ „
2(lnx) +51nx d(\nx) = -A L. 4. _A L. 4- Q.
-•^ ^ J 3 2
4. Quan he
giira
e^ va ae^ + b
Ta CO {^ae' = ae' nen quan he can xet giira va ae' + b la:
a
(a^O)
(ta
hiiu cong thuc tren mot each dan
gicin
nhu sau : dua 6 vao trong vi
phdn thi thanh
{ae""
+ h), voi a ^Ovdb tiiyytren M)
Vi
du 4.
Tinh
:
'
-J 2e' +1
b)
[—1—dx.
Gidi
a)Phan
tich
bdi
todn

: Cdc ban de y quan he giua va 2e^ +
1
.•
e'^dx
= ^ d(2e^ + 1) nen ta c6
dx - .e'dx =
.
- d(2e' + 1)
2e" +1 2e" + 1
3 1
2 2e' +1
2e^ + l 2
(i(2e^
+1). Do vgy ta chon dnphu la u
=
2e' +\.
Ld'i
gidi
cua bdi
todn
I
=
r-^-Hl-dx=
r-^.eMx=
.id(2e''+l)
1 J 2e^ +
1
^ 2e^ +1 ^ 2e'' +1 2 '
= r ^^d(2e^+l) = - f—^d(2e^+l).
2 20" +

1
2 26" + 1
Dat u = 26" +1 =^ du = d(2e'+1).
3 rl , 3
Ta CO
'2^11 2
In
u +C-
Thay u = 26" +
1
ta
dugc
: 1 = - ln(2e"' + 1) + C
•
(ta khong lay dau gia
tri
tuyet d6i vi 2e'' +1 > 0)
b) Phdn
tich
bdi
todn
: Ta Men doi
h
=
r
1_
l
+
e
-X

-dx =
1
1
+
1
-dx = -dx
Cdc
ban de y quan he
giiea
va e'^ +
1
e'^dx
= d(e* + 1) nen ta c6:
•
dx —
1
•
.e'^^dx
1
-d{e'
+!)•
+
1 + 1 + 1
Do
do, ta chon dn phu la u = e"
+
\.
LM
gidi
cua bdi

todn
1
l
+ e
-X
-dx =
—-dx = -dx =
1
+
-
e^+1
e'dx
e'+\
^
+ 1
Dat u = e''+l^du = d(e''+l).
Taco: I = f-du = ln|u|+C.
u
;
Thay u = e" +1 ta dugc: = ln(e" + 1) + C
*
Nhan xet: Neu da thanh thao trong viec su dung phuang phdp nay, cdc
ban
CO
the trinh bay lai gidi nhanh han nhu sau:
a) I = r—dx=
r—-— d(2e''+
l) = -ln(2e^+1) + C.
'
-'20^+1 2e^ +1 2 ^ ^ 2 ^ ^

b)
I2
=
l
+ e
-dx = -dx =
-dx
1
+
e^+1
=
f—^ d(e'' + 1) = ln(e^ + 1) + C
•
c'' + 1
5.
Quan he
giua
sinx
va cosx
Ta
CO
(sinx) =cosx va (cosx) =-sinx nen quan he can xet giua sin a;
va cos X la:
cos xdx =
—
d{a s inx+b)
a
s inxdx =
— —
d(a cos x + b)

a
(Ta
hieu cong thuc tren mot cdch dan gidn nhu sau: dim cos x vao trong
viphdn thanh (asinx+h); dua sinx vao trong vi phdn thanh -(acosx + b), vai
Vi
du 5 Tinh :
a)
=
Jcos^xsin'xdx;
b)
Jcosxe-'"""+'da:.
Gidi
a)
Phan tich bai todn : Ta bien doi :
cos^ X
sin'^
x= cos x cos^ x sin^ x= cos x{\ sin^ x) sin^ x .
Cdc ban de y quan he giua sinx va
cosx;
cosxdx = d(sinx) nen ta c6
cosx(l -sin^x)sin^xdx = (1 -sin^x)sin^xd(siwc). Do vdy, ta chon an phu la u = sinx.
LM
gidi cua bai todn
=
J cos^ x sin^
xdx=J
cos x cos^ x
siii^
xdx
=J

cos x(l - sin^ x) sin^ xdx = f{l- sm x)
sin^
xd(sin x).
Dat u = sinx => du = d(sinx).
Taco:
= J (1 -
u')uMu
= J (u'- u')du = y - y+0-
™
• . i T (sinx)^ (sinx)^ „
Thay u
=^
sinjc ta
duac:
I =-^^ ^ -^^ —+C-
' 3 5
b)
Phan tich bai todn : Cdc ban de y quan he giua sinx va
cosx;
cos xdx = d{-3 sin x + 2) nen ta c6
cos = —
e-'^'^'^^di-?,
sin x + 2)-
3
Do vdy, ta chon an phu Id u = -3sinx + 2.
LM
gidi cua bai todn
=
Jcosxe-'''"''^'dx =
J^e-"'"'^+'(i(-3sinx

+ 2).
Dat u = -3 sin x + 2 =^ du = d(-3 sin x + 2).
Taco:
I = — f eMu - —e" +C.
' 3 ^ 3
Thay u = -3 sin x + 2 ta
dugc:
= —
e-3»""'+2
^ Q.
* Nhan x4t: Neu da thanh
thgo
trong
viec
sir dung phucmg phdp nay, cdc
bgn
CO
thi trinh bay lai gidi nhanh han nhu sau :
a) Ij = J cos^ x sin^
xdx=J
cos x(l — sin^ x)
sin^
xdx
— J (I- sin^ x) sin^ xd(sin x) = J* (sin^ x - sin^ x)d(sin x)
sin^
X
sin^
x
b)/, = /
cos xe

1
-3
si
3 ^
3sinx+2
5
-3
sin
1+2
+c.
+
c.
dx^j
-1
-38mx+2
c/(-3sinx
+ 2)
6. Quan he gifra sin^x, cos^x va sin2x
Ta CO (sin^ x)
=2sinxcosx
= sin2x va (cos^ x)
=-2cosxsinx
=-sin2x
nen quan he can xet giua sin^x,
cos'^x
va sin2x la :
sin
2xdx =
—
d{a sin^ x -\-b)

a
sin
2xdx =
— —
d{a cos^ x + b)
a
(ta hieu cdng thuc tren mot cdch don gidn nhu sau: dua sin2x vao trong vi
phan thanh (a sin^ x + b)
hogc
—[a cos^ x + b), voi a ^0 va b
tuyy
tren M)
Vi
du 6. Tinh :
sin2x
a) = j(3siu^x+l)sin2x(ix; b) = r_^_sin2z
^ V2sin'rz;+•
rdX
•
3 cos X
Gidi
a) Phan tich bai todn : Cdc ban diyquan he giCta sin^x va sin2x;
sin2xrfa; =-d(3sin^ x+1) nen ta c6 (3sin^ x+l)sin2xda;
3
=
•^(3sin^ x+l)d(3sin^ x+1). Do vdy, ta chon dnphu la u = 3sin^x + I.
o
L&i
gidi cua bai todn :
= J(3sin'x+l)sin2xdx = J^(3sin^ x+l)d(3sin^ x+1)-

E>at u = 3sin^ x+1 ^ du = d(3sin^ x+1).
Taco
: I = - /
udu
= hC= hC-
Thay
u =
3sin'x+l tadugc
: I ^
(3sin
x+1)
'
6
b)
Phan
tick
bai
todn
:
Ta bien doi:
sin
2x sin 2x sin 2x
V2sin^
X
+
3cos^
x
-^2(sin^
x +
cos^ re)

+
cos^
x Vs +
cos^x
Cdc
ban
di y
quan
he
giua cos^x
va sin2x sin2xda; = -d{2 +
cos^
x) nen
ta
CO sin 2a:
_ =d{2 +
cos^ .x)-
do, ta c6 the
chon
an
V2
+
cos^a; V2
+
cos^x
/fl
w
= 2 +
cos^ X /zoac
w

=
V2 + cos^x. r/-o«g truang hap nay
ta
nen chon
u
= V2
+ cos" X
cfe
Z>/ew //zii-c i/j/OT i/aw nguyen ham khong
con can
thuc.
L&i
gidi
cua bdi
todn
:
r
s\n2x
, r —1 2 \
h=J
-/ , dx = J ,
c^(2
+
cos^
x)
•
^
V2sin^x
+
3cos^x

^
V2
+
cos^
x
Dat u
= V2 +
cos^
X => = 2 +
cos^
x
=> du^
= d(2 +
cos^
x).
Tac6:l^=r
jdidu^ =-J^du
=
-2jdu
= -2u + C-
Thay
u = y/l + cos' x
ta
dugc
: I2 = -2\/2 +
cos^
x + C.
*
Cach
khac

:
,
r
sm2x
, r -1 , ^
/2
= / , dx = i =rf(2 +
cos''x)-
^
V2sin^x
+
3cos^x ^l2
+
cos^x
Dat
u = 2 +
COS"
X
=^ du
= d(2 +
cos^
x).
Ta
CO : \ f
-r=du
=
—2Vu+C
(chu y
cong
thuc

dao ham
(V^)'=4=)-
2Vx
Thay
u
= 2 +
cos^x
ta dugc:
T = -2V2 +
cos^
x
4-
C•
i
•
*
Nhan
xet:
- A'ew thanh
thgo
trong viec
sir
dung phuang phdp nay,
cdc ban c6 the
trinh bay
lai
gidi nhanh han nhu sau
:
a)
=

J(Ssin^
x+l)sin2xc?x
=
j^(3sin^ x+l)c/(3sin^
x+1)
_(3sin^x+l)^
6
V2sin^
X
+
3cos^
x
yJ2
+
cos^
x
=
-
'(2
+
cos^x)
2
t/(2
+
cos^x)
=
-2V(2
+
cos^x)
+

C
-
A'ew
chung
ta de y
den quan
he
giua sin a;
va cos x thi
chung
ta c6
them each gidi theo huang khdc nhu sau
:
a) /j
=
J(3sin^
x+l)sin2x(ix
=
j2sinxcosx(3sin^ x4-l)(ix
=
J2sinx(3sin^ x+l)rf(sinx)=y
(6sin^
x+2sinx)(i(sinx)
„
sin^
X „
sin^
x „ 3 . 4 . , ^
=
6. h2. + C =

-sin^
x+sm^
x + C
•
4
2 2
Chii
y
rang each nay
va
each tren
deu cho ket qua
dung,
no chi sai
khdc
nhau
mot
hang so
xdc
dinh
!
b)
= f ^"^^^ dx =
flpi^dx
(sau do dua
sinx
^
V2sin^x
+
3cos^x

^
V2
+
cos^
x
vdo trong vi phan)
r
-2cosx
r 1 1/ 2 x
=
/
==rd(cosx)
= — I
—=====rd(cos
x)
^ V2
+
cos^x
^
V2
+
cos^x
-1
2
=
-
|(2
+
cos^x)
2

J(2
+
cos^x)
= _
(2±£^ij0i_
^
^
-^1
1
=
-2 (2 +
cos^x)2
+C = -2V(2 +
cos^x)
+ C.
1
1
7.
Quan
hf
giira
— va
tanx
,
cos^x
sin^
X
va
cotx
Ta

CO
(tanx) =—^-j-
va
(cotx)
= . ] nen
quan
he can xet
giua
—^
1
vd
tana;,
sin^
X
cos
X
vd cot
X la :
sin
jc
cos^x
^
dx =
—
d{a
tan
x+b)
a
cos^x
^

dx = — — d{a cot x+b)
a
sin^
x
(ta
hieu cong thuc tren
mot
each
dan
gidn
nhu sau : dua
1
cos^x
vdo trong
vi
phdn thdnh (atanx
+ b), dua
vai
a
^Ovdb tiiyy tren
R)
Vidu
7.
Tinh:
3 tan
X + 4
1
sin^
X
vdo trong

vi
phdn thdnh -(acotx
+ b),
a)/
+
cos
2a;
dx
\
cotx
smx
dx-
Gidi
a) Phan tick bdi todn
: Ta
bi6n
doi
3tana;
+ 4
3tana;
+ 4
(3tana;
+ 4) 1
1
4- cos
2x
2
cos
X
Cdc

ban de y
quan
he
giira
1
cos X
vd tan
X: dx = - d(3
tan
a;
+ 4)
2 2 o
ODS
a; cos a; o
mn,ac6 0tanx+
4) 1 ^
01'^"^
+
4)irf(3tanx
+ 4). Dov^.
2 cos X 2 3
ta
chon an
phu la M =
3 tan
x + 4.
LM
gidi cua bai todn
/
3tanx

+ 4^^ _ r
3tanx
+ 4
l
+ cos2x -J 2cos^x
dx
(3 tan
X + 4) 1
COS^
X
<lx
= /
(3tanx
+ 4) 1 ^ ,
-
d(3
tan
x + 4)
2 3
Dat
u =
3 tan
x + 4 du = d(3
tan
x + 4).
1
r 1
Ta CO
: T = - / udu = -
i^

+ c = -^ + c.
6 2 12
Thay
w =
3tanx
+
4tadugc:
T =
tanx+
4)^^^
'
12
b) tick bai todn
:
Ta
biin
doi
cotx
sinx
sin''
x
~
cot^
X.
—\
•
Cdc ban di y
quan
he
sin^

x
giica
—^
cotx;
-^dx =
-(i(cota;)
sin
X
sin
X
to
CO
cot'
a;.
—^
da;
= - cot'
xd(cot
a;)
•
-Do vdy ta c6 thi
chon dn
phu Id
sin'
X
M =
COtX.
L&i
gidi cua bdi todn
r

cot
X , r cot" x , r
I
dx = /
—^a;
= I
sm
X
cot
X
sm
X
cot'
X.
—^—dx
sin
X
=
-
J
cot'
a-(i(cot
a;).
Dat
u = cot
X
=> du =
d(cot
x).
Taco:

=
-J
u'du = -y+ C.
Thay
u =
cotx
ta
dugc:
I2 = - + C
•
3
* Nhgn
xet: Niu da
thanh
thgo
trong viec
sir
dung phuang phdp
nay, cdc
ban
CO
thi
trinh
bay lai
gidi nhanh
han nhu sau :
,s ^ f 3
tan
X + 4 . r 3
tan

x + 4
^1 = J T
1
r 3
tan
x
-dx
= I :
^X
+
cos 2x -^2 cos X
=
r(3tanx
+
4)
1
3^^^^^ (3tanx+4f
J 5! 3 ^ 12
(
^2
—
dx = I cot
X.
-—-—dx — — I cot xd(cot x)
sin X sin V ^
3
sm X
= 5—+ c-
Vay
la, chung ta da nghien

cihi
xong 7 m6i quan he ca ban giup chung ta
dinh
huang nhanh
each
giai
cho mot bai toan nguyen ham,
ciing
nhu
tich
phan
sau nay. Trong
truoiig
hop bai toan khong c6 xuat hien mot trong 7 moi quan
he tren, chung ta lam
theo
huang
giai
khac, c6
tinh
chat
tong quat hon nhu sau:
dat an phu u = u(x) de tit nguyen ham theo bien x chung ta bieu dien duac
nguyen ham do theo bien u!
(tiic
Id ta can biiu dien bien "x" theo bien u ,
"dx" theo u vd du)
Mb'i
cac ban
theo

doi mot so vi du minh hoa.
Vi
du 8.
Tinh
:
a) j x(x +
12f'Mx
; \
^'('^
+
l)'dx
•
Giai
a) Phan tich bdi toan : Niu khai
triin
(x +12)^'"^ rdi nhdn x vac di tinh thi
khong kha thi rdi ! O day chung ta
cUng
khong nhin
thdy
su xuat hien cua mot
trong
7 moi quan he de dinh huang phep dqt an phu, nhung theo huang giai
tong quat, chung ta chon an phu la u = x + 12 thi
tie
nguyen ham theo biin x
chung ta bieu dien duac nguyen ham do theo bien u rdi! Vi tit u = x + 12 ta
CO
X = u
—

12 va dx — d{u -12) — du (tuc la x duac biiu diin theo u vd
dx duac bieu dien theo du).
L&i
giai
cHa
bdi toan
Dat u = X + 12 =^ du = dx va x = u -12.
Ta
dugc
:
I
= r(u -
12)u^"^Mu
=
r(u^"i^
- 12u^"^^)du = - + C
'
' J ^ ' 2014 2013
Thay u ^
X
^ 12 ta
CO
: I ^ + '^f- _ 12(x + 12r ^ ^.
'
2014 2013
b)
Plian
tich bdi toan : Doi vai nguyen ham nay, vice su dimg moi quan he
giita
xvdx khong dem Igi lai giai thoa dang! Nhung neu chon an phu la u

=
x
1 thi tit nguyen ham theo biin x chung ta bieu dien duac nguyen ham do theo
biin
u rdi! Vi tit u = x + \ c6 x = u - \ dx = d(u - 1) = du
(tiic
Id x duac
••
biiu diin theo u vd dx duac bieu dien theo du).
\ giai cua bdi toan
Dat u = x+
1
=>du = dxvax = u- l.
Ta
dugc
:
= J(u -
l)'uMu
= J(u' - 2u +
l)uMu
- /(u' -
2u'*
+ u')du
2n'
u u
10 9
Thay u = x +
1
ta c6 :
(x

+ ir 2(x +
l)°
^ (x + 1)^
10 9 8
1^
=
+
C.
Vi
du 9.
Tinh
:
+1
{x-2)
-dx.
Giai
a) Phan tich bdi toan :
Trong
bdi nay
cUng
vay,
su dung moi quan he giita x
vd
x^ khong dem Igi lai giai thoa dang! Neu chon an phu la u = x - 2 thi tit
nguyen ham theo biin x chung ta bieu dien duac nguyen ham do theo bien u!
Vi
tit u =
X
- 2 ta
CO

X
= u + 2 vd dx = d(u + 2) = du (tuc la x duac bieu dien
theo u vd dx duac bieu dien theo du).
Lcfi
giai cua bdi toan
Dat u = x- 2=>du = dxvax = u + 2.
Ta
dugc
:
r
(u + 2)' +
1
, r u' + 4u + 5 , r, I , 4 , 5 , ,
-9 4 -10
r,
.,M . 11 . ^ , u 4u
-1
-10
+
4u-'^+5u^^)du = —+
+
9 -10 -11
+
C
9u^
lOn'' ' iW
Thay u ~ x - 2 ta c6 :
I.
=
-1

-4
-5
'
9{x-2f
10(x-2r
ll(x-2)"
*
Chuy
: Neu dp
dung cong thuc
+
C-
,
chung
ta
CO
cdch gidi
gon
han!
(u
+
2)^
+ l
I
1
J
U
-1
-4
+

:r +
-5
+
4u + 5
.12
du
T/
1 4 5 ,,
9u'
lOu^" llu^^
Thay
u = x - 2 ta c6 :
+
C.
1.
=
-5
'
9(x-2)' 10(x-2y"
ll(x-2)"
+
C.
b)
Phan
tich bai
todn
:
Ta
cd
bien

doi
X
-
.x^
nen
chung
(2
+
x^f
(2 +
x^)^
ta
CO
the su
dung
moi
quan
he
giua
x^ vd x^ de
dinh huang phep
dqt
an
phu:
xMx
= -d(2 + x').
3
Ta
CO
-dx

-
(2
+
x^)'
{2
+ xy
chon an
phu la u
=
2 +
x\
LM
gidi
cm
bai
todn
1
X
.x'dx
= —
d(2
+ x'), vgy ta
3(2
+
x')'
it
u =:
Ta
dugc
:

.xMx=
r •-d(2 + x^)
'
'(2 +
x^)^^^^
^ (2 +
x')^
Dat
u = 2 + x^ du = d(2 +
x^)
va x' = u - 2.
3 u 3 u u 3 u u
=
i(ln|u|
+ ^) + C.
3 u
Thay
w
=
2 + xMac6
: I =-(In
'
3
2
+ x'
+
2
+ x'
r)
+ C-

Vi
du 10.
Tinh
:
(sinx
+
cosx)
a) I, = f
_
dx; b) I = r sin^ xVcos xdx.
Gidi
a)
Phan
tich bai
todn
:
Ta
bien
doi
cos X cos
X
(sinx
4-cosx)^ ,sina; + cosx., 3 (tanx + 1)^ cos^ x
^
' ( ) cos X '
cos
X
Tai
ddy
chung

ta cd the su
dung quan
he
giua
—^— vd
tanx
de
chon
an
cos^ X
phu thich
hap la u
=
tanx
+
1.
LM
gidi
cua
bai
todn
Ta
CO
:
r
cosx
, r
cosx
,
Ii

= / — = I — dx
(sinx
+
COSx)^
^ ,sinx +
cosx.3
3
^
^ ( ) COS X
cosx
=
f -—i—dx= r -d(tanx + l)
J
(tan X +
1)^
cos^ x <^ (tan x +
1)^
'
Dat
u = tan x +
1
du = d(tan x + 1).
Tadirgc:
T ^ T-^du = f
u-Mu
- — + C - ^ + C•
^
J J -9.
-2
2u'

Thay
u = tan x +
1
ta
c6:
i = —
1-
C
•
'
2(tanx
+
lf
*
Nhan
xet:
-
Trong
bai
todn nay, phdi thong
qua mot
sSphep bien doi, chung
ta mai dp
dung duac quan
h$
giua
\— vd
tanx
di
dinh huang phep

dat
an
phu
cUng
COS^
X
nhu cdch gidi!
-
Cdc ban cd thi
trinh
bay lai
gidi
gon han, vai chu y
cong thuc
I
=
r——^ix
= f—
(sinx
+
cosx)
^ ,smx
cos X
4-cosx,3
3
f
cos*
x
-dx
cosx

=
f—-
1
(taiix
+ 1)^ cos^ X
-1
•dx
(tan X + 1)"
•d(tan
X
+ 1)
+
C-
2(tan
X
+ 1)^
b)
Phan
tick
bai
todn
: Ta de y quan he giita sin x vd cos x de dinh
huangphep dgt an phu : sin xdx =
—d(cos
x). Ta cd
sin^
xVcosxdx
= sin x sin^
xVcosxdx
=

sin x(l
—
cos^
x)Vcosxdx
=
(1 - cos^
x)Vcos
xd(— cos x)
vgy ta chon an phu la u
=
cosx
hoac
u
=
Vcosx
.
Trong
truang hap nay
chung ta nen chon u
=
Vcosx
de bieu thuc duai ddu nguyen hdm khong con
chica
can thuc.
Ldi
giai
cua bai
todn
\
J

sin^
xV
cos xdx =
J
sin x sin^
xVcosxdx
=
sin
x(l
- cos^
x)
V
cos xdx
=
(1
- cos" x)Vcosxc/(-cosx)
Dat u = VcosX =4> =
cosx
va
d(—cosx)
= d(—u^).
Tadugc: = J(1 -
u^)ud(-u')
= J(1 -
u')u(-2u)du
=
r(-2u^+2u«)du
= Z^+2u^ + C.
^
3 7

Thay u - ta c6 : =
-2{sf^f
^ 2(V^)^
*
Cach
khac : Neu dat
M
= cos
x
ta c6
each
giai
khac
nhu sau :
\—j'^ii^^
xVcosxdx
—
J'sinXsin^
xVcosxdx
= j sin x(l - cos^
x)Vcosxdx
= J{1- cos^
x)Vcosxd(-
cos x)
DSt u = cos X du —
d(cos
x).
Tadugc : = \(l-u')^f7^d{-u)^ \(\-u'yd{-u)
=
-j

u^-u 2
du
/
5 J^>|
-u^
\du-
7 3
2w^
Thay
w
= cos x ta c6 :
2^Vcosxj
2(Vcosx
j
_2(cosx)2 2(cosx)2
_^
'
7 3 7 3 •
Cdc
ban nhdn
thdy,
bdng cdch dat u =
Vcosx
chung ta c6 lai gidi gon han
vd
khong phuc tap nhu cdch dat u
=
cosx.
Vidu
11.

Tinh:
a) I, =
rdX
'
b) = J
xVl
+ xdx.
Gidi
a) Phan
tick
bai
todn
: Bai todn ndy sit dung 7 quan he de dinh huang phep
dat dn phu Id khong khd thi. Neu chon an phu Id u = Vl
—
x ihi tit nguyen
hdm theo biin x chung ta bieu diin duac nguyen hdm do theo bien u vi tit
u
= Vl -
X
ta suy ra x =
1
- vd dx = d(l - u^) = -2udu (tuc Id X
duac bieu dien theo u; dx duac bieu dien theo u vd du ).
L&i
gidi
cHa
bai
todn
Dat Li = Vl -X =^ =

1
- X hay X =
1
- dx = d(l - u^)
=
-2udu-
Ta
duac
gc : =
-Ji-iil2udu
= 2J(u' - l)du = 2
u
u
+
c
(Vr^)^
+
c-
Thay
u = Vl -x ta c6: = 2
b)
Tuang tu cdu a), ta chon dn phu la u ^ Vl + x. Lai gidi cua bdi todn
Id
Dat u = Vl + x = 1 -f
X
hay X = -1 dx = d(u^ -1)
=
2udu
u
—

l)u.2udu
=.2/(u^-u=)du
= 2(^-^) + C
Thay
+
c.
Vi
du 12.
Tinh
:
dx
e" +1
b)
l, = J
„2x „x
e
—
e
dx.
Gidi
a) P/tan tick bai todn : Neu
chon
an phu la
M
= + 1 thi tir nguyen ham
theo
bien
x chung ta
bieu
dien

duac
nguyen ham do
theo
bien
u vi
du — rf(e^ +1) = e'^da: ma
=-u-\.
LM
gidi cua bai todn
Dat
u = + 1 <^ = u -1 =^ du = d(e'' + 1) = eMx
du
du
dx
= — =
u-1
Ta
duac:
=
ln
du
(u
- 1)
u-1
'
J nin-l) J
u-1
u
du
= In

u-1
-In
u
+
C
u
+
C.
Thay
u = +1 ta c6: ~ in
e^+1
-)-
C
•
(ta khong lay ddu gid tri
tuyet doi vi
+1
b) Pit
an tick bai todn : Chu y rdng e^'' = (e'')^ tuc Id e^" biiu diin duac
de
qua e"". Lai de y de" = e^dx hay = dx nen ta
chon
dn phu la u = e\
L&i
gidi cua bai todn
E)5t u ^e'' =^du = de''.
r
1 du r 1 , pn^ ~ (u"
Tadugc:
= / — •— = / -du =J

^
u^ - u u u (u -1) ^ u (u -
u^
-
(u^
-1)
1)
du
-I
=
In
u
(u^
-1)
[u^(u-l)
u'(u-l)J
1
1 1
du
u-1
u u
du
= In
-I
u-1
1
u + 1
u-1
u'
1

du
-In
u
+
- + C
u
u-1
u
+
i
+
c
u
Thay
u = e'' ta c6 : hi
e" -1
*
Cdch khdc : Cdc ban diyrdng
1
1
_ je-f
vd
d{e"'') = -e~\ix hay ^—^ = dx thi ta
chon
duac
dn phu la u — e
Ta
CO
lai gidi cho bai todn:
=

r_J_^ix=
f—^—dx=
fi^
dx
=
I
-^^^x
du
Dat
u = e " du = de " = -e ''dx hay — = dx
•
-u
Tadugc:
1= - f'-^idu = f —
•^1-uu
-'u-
Thay
u = e"" ta c6 :
l-l
du
= u + hi
u-1
u
- 1 + C .
I,
= e-'' + In
e"" -1
+
C = In
e" -1