Giáo trình giải tích 1 part 6 ppt

F f (a, b)

f(x)dx = F (x)+C
C


f(x)dx


= f(x)

F

(x)dx = F (x)
f(x)=(x sin
1
x
)

f(x)= (x)
f,g α, β ∈ R

(αf(x)+βg(x))dx = α

f(x)dx + β

g(x)dx
x = ϕ(t) J
f(x) I = ϕ(J)

f(x)dx =


f(ϕ(t))ϕ

(t)dt =

f(ϕ(t))dϕ(t)
u, v

u(x)v

(x)dx = u(x)v(x) −

v(x)u

(x)dx

udv = uv −

vdu

x
C

x
α
dx =
x
α+1
α +1
+ C (α = −1)


1
x
dx =ln|x| + C

a
x
dx =
a
x
ln a
+ C

e
x
dx = e
x
+ C

sin xdx = −cos x + C

cos xdx =sinx + C

1
cos
2
x
dx =tanx + C

1

sin
2
x
dx = − x + C

dx
x
2
+ a
2
=
1
a
arctan
x
a
+ C

dx
x
2
− a
2
=
1
2a
ln





x + a
x − a




+ C

dx
√
a
2
− x
2
=arcsin
x
a
+ C

dx
√
x
2
± a
2
=ln|x +
√
x
2

± a
2
| + C


a
2
− x
2
dx =
x
2

a
2
− x
2
+
a
2
2
arcsin
x
a
+ C


x
2
± a

2
dx =
x
2

x
2
± a
2
±
a
2
2
ln |x +

x
2
± a
2
| + C

(2
x
+sinx −
1
3
√
x
)dx =


2
x
dx +

sin xdx −

x
−
1
3
dx =
2
x
ln 2
−cos x−
3
2
x
2
3
+ C

dx
x
2
+ a
2
=
1
a

2

dx

x
a

2
+1
t =
x
a
dx = adt

dx
x
2
+ a
2
=
1
a

dt
t
2
+1
=
1
a

arctan t + C =
1
a
arctan
x
a
+ C


a
2
− x
2
dx x = a sin t, t ∈ [−
π
2
,
π
2
]
dx = a cos tdt


a
2
− x
2
dx = a
2



1 − sin
2
t cos tdt = a
2

cos
2
tdt
= a
2

cos 2t +1
2
dt =
a
2
2
(
sin 2t
2
+ t)+C =
a
2
2
(sin t cos t + t)+C
t =arcsin
x
|a|



a
2
− x
2
dx =
x
2

a
2
− x
2
+
a
2
2
arcsin
x
a
+ C


x
2
+ a
2
dx x = a sinh t = a
e
x

− e
−x
2
dx = a cosh tdt x
2
+ a
2
= a
2
(sinh
2
t +1)=a
2
cosh
2
t


x
2
+ a
2
dx = a
2

cosh
2
tdt = a
2



cosh 2t +1
2

dt
=
a
2
4
(sinh 2t +2t)+C =
a
2
4
(2 sinh t cosh t +2t)+C
e
2t
−
2x
a
e
t
− 1=0 t =ln

x +
√
x
2
+ a
2
a




x
2
+ a
2
dx =
x
2

x
2
+ a
2
+
a
2
2
ln
x
+

x
2
+ a
2
| + C

dx

√
a
2
− x
2

dx
√
x
2
± a
2

f
α
(x)f

(x)dx

x
2

x
3
+5=

(x
3
+5)
1

2
d(x
3
+5)
3
=
1
3
2
3
(x
3
+5)
3
2
+ C

sin
4
x cos xdx =

sin
4
xd(sin x)=
sin
5
x
5
+ C


tan xdx =

sin x
cos x
dx = −

d(cos x)
cos x
= −ln |cos x|+ C

(ax + b)
α
dx

cos
3
x sin xdx

xdx

P (x)lnxdx

P (x)e
ax
dx

P (x)sinaxdx

P (x)cosaxdx
P

I
n
=

x
n
ln xdx
n = −1
u =lnx ⇒ du =
dx
x
dv = x
n
dx v =
x
n+1
n +1
I
n
=
x
n+1
n +1
ln x −
1
n +1

x
n
dx =

x
n+1
n +1
ln x −
x
n+1
(n +1)
2
+ C
n = −1 I
−1
=

ln x
x
dx =

ln xd(ln x)=
ln
2
x
2
+ C
I =

(x
2
+ x +1)sinxdx
u = x
2

+ x +1 ⇒ du =(2x +1)dx
dv =sinxdx v = −cos x
I = −(x
2
+ x +1)cosx +

(2x +1)sinxdx
u =2x +1 ⇒ du =2dx
dv =cosxdx v =sinx

(2x+1)sinxdx =(2x+1)sinx−2

sin xdx =(2x+1)sinx+2cosxdx+C
I = −(x
2
+ x +3)cosx +(2x +1)sinx + C
A =

e
ax
cos bxdx, B =

e
ax
sin bxdx
dv = e
ax
dx
A =
1

a
e
ax
cos bx +
b
a

e
ax
sin bxdx =
1
a
e
ax
cos bx +
b
a
B
B =
1
a
e
ax
sin bx −
b
a

e
ax
cos bxdx =

1
a
e
ax
sin bx −
b
a
A
A =

e
ax
cos bxdx =
b sin bx + a cos bx
a
2
+ b
2
e
ax
+ C
B =

e
ax
sin xdx =
a sin bx −b cos bx
a
2
+ b

2
e
ax
+ C

P (x)sinaxdx = A(x)sinax + B(x)cosax + C A, B
< P A, B
ln x, arctan x, arcsin x
I
n
= I
n
(a)=

dx
(x
2
+ a
2
)
n
(n ∈ N)
I
1
=

dx
x
2
+ a

2
=
1
a
arctan
x
a
+ C
n>1
I
n
=
1
a
2

x
2
+ a
2
(x
2
+ a
2
)
n
dx −
1
a
2


x.x
(x
2
+ a
2
)
n
dx
=
1
a
2
I
n−1
−
1
a
2

−
x
2(n − 1)(x
2
+ a
2
)
n−1
+
1

2(n − 1)
I
n−1

I
n
=
1
2a
2
(n − 1)
x
(x
2
+ a
2
)
n−1
−
2n − 3
2a
2
(n − 1)
I
n−1
•
P (x)
Q(x)
P (x)
Q(x)

= M(x)+
P
1
(x)
Q(x)
M(x) P
1
(x) < Q(x)
Q(x)=A(x − a)
m
···(x
2
+ px + q)
n
···
a Q p, q p
2
− 4q<0
P
1
(x)
Q(x)
=
A
1
x − a
+ ···+
A
m
(x − a)

m
+ ···
+
B
1
x + C
1
x
2
+ px + q
+ ···+
B
n
x + C
n
(x
2
+ px + q)
n
+ ···
A
i
,B
i
,C
i
1
x − a
1
(x − a)

m
Bx + C
x
2
+ px + q
Bx + C
(x
2
+ px + q)
n
(p
2
− 4q<0)

P (x)
Q(x)
dx =

M(x)dx +

P
1
(x)
Q(x)
dx

M(x)dx

dx
x − a

=ln|x − a| + c

dx
(x − a)
m
=

d(x − a)
(x − a)
m
=
1
(1 − m)(x −a)
m−1
+ c (m =1)

Bx + C
x
2
+ px + q
dx =
B
2

d(x
2
+ px + q)
x
2
+ px + q

+(C −
Bp
2
)

dx
x
2
+ px + q
x
2
+ px + q =(x +
p
2
)
2
+
4q −p
2
4
t = x +
p
2
a =

4q −p
2
2

Bx + C

x
2
+ px + q
dx =
B
2
ln |x
2
+ px + q| +
2C − Bp

4q −p
2
arctan
2x + p

4q −p
2
+ c

Bx + C
(x
2
+ px + q)
n
dx =
B
2

d(x

2
+ px + q)
(x
2
+ px + q)
n
+(C −
Bp
2
)

dx
(x
2
+ px + q)
n
t = x +
p
2
a =

4q −p
2
2
I
n
=
1
2a
2

(n − 1)
x
(x
2
+ a
2
)
n−1
−
2n − 3
2a
2
(n − 1)
I
n−1
= ···
=
<n− 1
(x
2
+ px + q)
n−1
+ A arctan

2x + p

4q −p
2

+ c



x
3
+ x +1
x
3
+ x
dx
x
3
+ x +1
x
3
+ x
=1+
1
x
3
+ x
x
3
+ x = x(x
2
+1)
1
x
3
+ x
=

A
x
+
Bx + C
x
2
+1
A, B, C
1 ≡ A(x
2
+1)+(Bx + C)x
1 ≡ (A + B)x
2
+ Cx + A
1,x,x
2
, ···
A =1,C =0,A+ B =0 ⇔ A =1,B = −1,C =0
1
x
3
+ x
=
1
x
−
x
x
2
+1


x
3
+ x +1
x
3
+ x
dx =

dx +

1
x
dx −

xdx
x
2
+1
= x +ln|x|−
1
2

d(x
2
+1)
x
2
+1
= x +ln|x|−

1
2
ln(x
2
+1)+C

dx
x
5
− x
2
x
5
− x
2
= x
2
(x − 1)(x
2
+ x +1)
1
x
5
− x
2
=
A
x
+
B

x
2
+
C
x − 1
+
Dx + E
x
2
+ x +1
1
x
5
− x
2
=
0
x
−
1
x
2
+
1
3(x −1)
−
x − 1
3(x
2
+ x +1)


dx
x
5
− x
2
=
1
x
+
1
6
ln
(x − 1)
2
x
2
+ x +1
+
1
√
3
arctan
2x +1
√
3
+ C

dx
x

4
− x
2
− 2

(x +1)dx
x
4
− x
2
− 2

x
2
dx
x
6
− 1

dx
x(x
2
+1)
2

(x −1)dx
(x
2
+ x +1)
2


(x
5
+1)dx
x
4
− 8x
2
+16
Q(x)=A(x − a)
n
···(x
2
+ px + q)
m
··· ,
Q
1
(x)=A(x −a)
n−1
···(x
2
+ px + q)
m−1
··· ,
D(x)=(x − a) ···(x
2
+ px + q) ···
P (x) deg P<deg Q


P (x)
Q(x)
dx =
M(x)
Q
1
(x)
+

N(x)
D(x)
dx
M(x),N(x) deg M<deg Q
1
, deg N<deg D
A, B, C, D, E

xdx
(x − 1)
2
(x +1)
3
=
Ax
2
+ Bx + C
(x − 1)(x +1)
2
+


(
D
x − 1
+
E
x +1
)dx
•

R(x,

ax + b
cx + d

r
1
, ··· ,

ax + b
cx + d

r
n
)dx R r
1
, ··· ,r
n
∈
Q
t

m
=
ax + b
cx + d
m r
i

dx
4
√
x +3− 1)
√
x +3
t
4
= x +3 dx =4t
3
dt

t
3
dt
(t − 1)t
2
=

tdt
t − 1
=4(t +ln|t − 1|)+C =4(
4

√
x +3+ln|
4
√
x +3− 1|)+C

dx
x(1 + 2
√
x +
3
√
x)
,

1 −
√
x +1
1+
3
√
x +1
dx ,

x

x − 2
x +1
dx


R(x,

ax
2
+ bx + c)dx R
ax
2
+bx+c a>0 t =
√
ax+
√
ax
2
+ bx + c
ax
2
+ bx +c = a(x −x
1
)(x −x
2
) t(x −x
1
)=

a(x − x
1
)(x − x
2
)


dx
√
x
2
+ bx + c
t = x +
√
x
2
+ bx + c
bx + c = t
2
− 2tx, bdx =2tdt − 2tdx − 2xdt,
dx
t − x
=
2dt
b +2t

dx
√
x
2
+ bx + c
=

dt
b
2
+ t

=ln




b
2
+ x +

x
2
+ bx + c




+ C

dx
(x
2
+ a
2
)
√
a
2
− x
2
t(a − x)=

√
a
2
− x
2
x =
a(t
2
− 1)
t
2
+1
dx =
4atdt
(t
2
+1)
2
1
2a
2

2t
2
+2
t
4
+1
dt =
1

2a
2


1
t
2
+
√
2t +1
+
1
t
2
−
√
2t +1

dt
=
1
a
2
√
2
(arctan(
√
2t + 1) + arctan(
√
2t − 1)) + C

t =

a + x
a − x

dx
x
√
x
2
+ a
2
,

dx
x +
√
x
2
+2x
,


−x
2
+4x +10dx
ax
2
+ bx + c = a


x +
b
2a

2
+

c −
b
2
4a

u = x +
b
2a
,du= dx

R(u,

α
2
− u
2
)du t = α sin u

R(u,

α
2
+ u

2
)du t = α tan u

R(u,

u
2
− α
2
)du t =
α
sin u

dx
(
√
a
2
− x
2
)
3
x = a sin t dx = a cos tdt

dx
(
√
a
2
− x

2
)
3
=

a cos tdt
(

a
2
− a
2
sin
2
t)
3
=

a cos tdt
a
3
cos
3
t
=
1
a
2
tan t+C =
1

a
2
x
√
a
2
− x
2
+C
•

R(sin x, cos x)dx R
t =tan
x
2

dx
1+ cos x
(0 <<1) t =tan
x
2
x = 2 arctan t, dx =
2dt
1+t
2
, cos x =
1 − t
2
1+t
2


dx
1+ cos x
=

2dt
(1 − )t
2
+1+
=
2
1 − 

dt
t
2
+
1+
1−
=
2
1 − 

1+
1 − 
arctan t

1+
1 − 
+ C =

2
1 − 
2
arctan


tan
x
2

1+
1 − 


+ C
R(−sin x, cos x)=−R(sin x, cos x) t =cosx
R(sin x, −cos x)=−R(sin x, cos x) t =sinx
R(−sin x, −cos x)=R(sin x, cos x) t =tanx

dx
2sinx − cos x +5

sin
3
x
2+cosx
dx

dx
sin

4
x cos
3
x

sin
m
x cos
n
xdx
m n t =cosx t =sinx
m, n

sin
4
x cos
5
xdx

sin
2
x cos
4
xdx
···

e
−x
2
dx,


sin x
x
dx,

cos x
x
dx

sin x
2
dx,

cos x
2
dx

x
m
(ax
n
+ b)
p
dx p,
m +1
n
,
m +1
n
+ p ∈ Z


dx

(1 − x
2
)(1 − k
2
x
2
)
,

x
2
dx

(1 − k
2
x
2
)(1 − x
2
)
,

dx
(1 + hx)
√
1 − k
2

x
2
0 <k<1
[a, b] P = {x
0
, ··· ,x
n
}
a = x
0
<x
1
< ···<x
n
= b
∆x
i
= x
i
− x
i−1
|P | =max{∆x
i
:0≤ i ≤ n}
f :[a, b] → R P
m
i
=inf{f(x):x
i−1
≤ x ≤ x

i
},M
i
=sup{f(x):x
i−1
≤ x ≤ x
i
}
L(f,P)=
n

i=1
m
i
∆x
i
U(f,P)=
n

i=1
M
i
∆x
i
✲
x
✻
y
abx
i−1

x
i
m
M
P, P

P
∗
= P ∪ P

P, P

I
∗
P
∗
I,I

P, P

I
∗
⊂ I,I
∗
⊂ I

inf
I
f(x) ≤ inf
I

∗
f(x) ≤ sup
I
∗
f(x) ≤ sup
I

f(x)
L(f,P) ≤ L(f,P
∗
) ≤ U(f, P
∗
) ≤ U (f, P

)
sup, inf
I
(f)=sup
P
L(f,P) I(f)=inf
P
U(f,P)
L(f,P) ≤ I
(f) ≤ I(f) ≤ U(f,P

) P, P

f
[a, b] f ∈R[a, b]
I

(f)=I(f ) f [a, b]

b
a
f

b
a
f(x)dx
f [a, b]
>0 P [a, b] U(f,P) −L(f,P) <.
f [a, b]
P
n
[a, b] U (f, P
n
) − L(f,P
n
) → 0. n →∞

b
a
f = lim
n→∞
U(f,P
n
) = lim
n→∞
L(f,P
n

)
f ≡ c f

b
a
f = c(b − a)
f [a, b] P = {x
0
, ··· ,x
n
}
[a, b] f(x)=c
i
x ∈ [x
i−1
,x
i
] f

b
a
f =
n

i=1
c
i
(x
i
− x

i−1
)
[0, 1]
D(x)=

0 x
1 x
P L(D,P)=0,U(D ,P)=1.
f :[a, b] → R
P = {x
0
, ··· ,x
n
} [a, b] ξ
P
= {c
1
, ··· ,c
n
} x
i−1
≤ c
i
≤ x
i
S(f, P,ξ
P
)=
n


i=1
f(c
i
)∆x
i
L(f,P) ≤ S(f, P,ξ
P
) ≤ U(f, P)
f [a, b]
lim
|P |→0
S(f, P,ξ
P
)=I
∀>0, ∃δ>0:∀P, |P | <δ ⇒|S(f, P,ξ
P
) −I| <, ∀ξ
P

b
a
f = I
• ⇒ f ∈R[a, b] >0 P
0
U(f,P
0
) <

b
a

f +

4
M =sup{f (x):a ≤ x ≤ b} n
0
P
0
δ
1
= min(/4Mn
0
, |P
0
|) P = {x
i
: i ∈ I}
|P | <δ
1
U(f,P)=

i∈I
M
i
∆x
i
=

i∈I
1
M

i
∆x
i
+

i∈I
2
M
i
∆x
i
,
I
1
= {i ∈ I :[x
i−1
,x
i
] P
0
}
I
2
= {i ∈ I :[x
i−1
,x
i
] P
0
}

δ
1
[x
i−1
,x
i
] P
0

i∈I
2
M
i
∆x
i
≤

i∈I
1
Mδ
1
≤ n
0
Mδ
1
≤

4

i∈I

1
M
i
∆x
i
≤ U(f, P
0
)+

i∈I
2
Mδ
1
≤ U(f, P
0
)+

4
( )
U(f,P) ≤ U(f,P
0
)+

2
<

b
a
f + 
δ

2
> 0 P |P | <δ
2
L(f,P) >

b
a
f −
δ = min(δ
1
,δ
2
) P |P | <δ ξ
P

b
a
f − <L(f,P) ≤ S(f, P,ξ
P
) ≤ U(f, P) <

b
a
f + 
• ⇒ lim
|P |→0
S(f, P,ξ
P
)=I >0 δ>0
P |P | <δ ξ

P
I −

2
<S(f,P,ξ
p
) <I+

2
P ξ
P
L(f,P)=inf
ξ
P
S(f, P,ξ
P
) U(f,P)=sup
ξ
P
S(f, P,ξ
P
)
I −

2
≤ L(f, P) ≤ U(f, P) ≤ I +

2