MËT SÈ K THUT "PH N TCH " M TA TH×ÍNG GP KHI I TMNGUYN HM HOC TNH TCH PH N.
Th½ dö 1 : T¼m nguy¶n h m A1 =
Z
dx(x + 1) (x + 2)
1
x + 1 =
1
−2 + 1 = −1C¡ch 2 :
2α +
1
3β =
16
⇔
(
α = 1
β = −1C¡ch 3 :
x (α + β) + 2α + β(x + 1) (x + 2)C¥n b¬ng c¡c h» sè 2 v¸ m ta câ h» :
(
α + β = 02α + β = 1 ⇔
Z
dx(x + 1) (x + 2) =
x + 1
x + 2
+ cTh½ dö 2 : T¼m nguy¶n h m A2 =
0 + 2(0 − 2) (0 + 5) = −
15
Trang 2α − β +1
6χ = −
121
3α + β +
1
8χ =
524
χ = − 335C¡ch 3 :
β = 27
χ = −335
Do â : x + 2
x (x − 2) (x + 5) = −
15x +
2
7 (x − 2) − 3
35 (x + 5)Vªy :
Z
x + 2
x (x − 2) (x + 5)dx = −
15
Z
dx
x +
27
Z
x2(−3x2− 2x + 5) (x + 1)dx
23
x + 53
= 14C¡ch 2 :
Trang 3β = −2548
χ = 14C¡ch 3 :
(x + 1)
=α
x +53
(x + 1) + β (x − 1) (x + 1) + χ (x − 1)
x +53
(x − 1)
x +53
3α +
2
3χ = 05
χ = 14
Z
x2(−3x2− 2x + 5) (x + 1)dx = −
116
+ 14
x +53
+ 1
4ln |x + 4| + cTh½ dö 4 : T¼m nguy¶n h m A4 =
Z
x − 1(x2+ 4x + 5) (x2− 4)dx
Ta c¦n chó þ r¬ng ph÷ìng tr¼nh : ax2+ bx + c = 0 vîi ∆ = b2− 4ac < 0 th¼ ta vi¸t :
ax2+ bx + c = a (x − x1) (x − x2)trong â : x1 = α + βi, x2 = α − βi, i2 = −1
x − 1(x + 2 − i) (x2− 4) = −
13
34 − 1
34i
Trang 4β = lim
x→−2+i
(x − 1) (x + 2 − i)(x + 2 + i) (x + 2 − i) (x2− 4) =x→−2+ilim
x − 1(x + 2 + i) (x2− 4) = −
x − 1(x2+ 4x + 5) (x + 2) =
168
δ = lim
x→−2
(x − 1) (x + 2)(x2+ 4x + 5) (x − 2) (x + 2) = limx→−2
x − 1(x2+ 4x + 5) (x − 2) =
34
x − 2+
34
χ = 168
δ = 34
β = −2717
χ = 168
δ = 34
Trang 5x − 1(x2+ 4x + 5) (x2− 4)dx = −
1334
Z 2x + 54
13
x2+ 4x + 5dx +
168
Z
dx
x − 2 +
34
Z
dx
x2+ 4x + 5 +
168
Z
dx
x − 2+
34
168
Z
dx
x − 2+
34
x − 1
2+
√3
2 i
= 1
6−
√3
x − 1
2−
√3
2 i
= 1
6 +
√3
6 i
x − 1
2−
√3
2 i+
1
6+
√3
6 i
x − 1
2+
√3
2α + β + χ =
121
χ = 13Vªy : x
Trang 6χ = 13Vªy :
Z
x
x3+ 1dx = −
13
Z
dx
x + 1 +
16
Z
2x + 2
x2− x + 1dx
= −13
Z
dx
x + 1 +
16
Z
2x + 1
x2− x + 1dx +
16
Z
dx
x2− x + 1
= −13
Z
dx
x + 1 +
16
Z d x2− x + 1
x2− x + 1 +
16
Z
dx
x −12
2+ 34
Z
dx
x4+ 1Gi£ sû r¬ng : x4+ 1 = x2+ px + 1 x2− px + 1= x4+ 2 − p2x + 1
2 i
x +
√2
2 −
√2
2 i
x −
√2
2 +
√2
2 i
x −
√2
2 −
√2
2 iT¼m c¡c h» sè : α, β, χ, δ nh÷ sau :
2 −
√2
8 +
√2
2 +
√2
8 −
√2
2 −
√2
8 +
√2
2 +
√2
8 −
√2
8 i
Vªy : 1
x4+ 1 =
√2
8 +
√2
8 i
x +
√2
2 −
√2
2 i
+
√2
8 −
√2
8 i
x +
√2
2 +
√2
2 i
x2+√
Trang 78 i
x −
√2
2 −
√2
8 −
√2
8 i
x −
√2
2 +
√2
x2+√
2x + 1 +
−
√2
4 x +
12
x2−√2x + 1C¡ch 2 : Ta gi£ sû r¬ng : 1
√2α + β+ √
5
√2α + β+ √
√2α + β + 5√
√2α − 4β + 5√
β = δ = 1
2
χ = −
√24
Vªy ta câ ÷ñc : 1
x4+ 1 =
√2
4 x +
12
x2+√
2x + 1 +
−
√2
4 x +
12
x2−√2x + 1C¡ch 3 :
β = δ = 1
2
χ = −
√24
Trang 8Do â : 1
x4+ 1 =
√2
4 x +
12
x2+√
2x + 1 +
−
√2
4 x +
12
x2−√2x + 1Vªy ta câ :
x2+√
2x + 1dx =
1
4√2
Z
dx
x + √12
2+ 12
= 1
4√
2ln x
2+√2x + 1+ 1
2√
2arctan
√2x + 1+ c1
x2−√2x + 1dx =
1
4√2
Z −2x +√2
x2−√2x + 1dx +
14
Z
dx
x −√12
2+12
2√
2arctan
√2x − 1+ cTh½ dö 7 : T¼m nguy¶n h m A7 =
Z
dx
x8+ 1Gi£ sû r¬ng : x8
p
2 −√2x + 2
x2+p2 −√
2x + 1
+18
p
2 +√2x + 2
Trang 9p
4ac − b2 2a
2 = 1a
= √ 2
4ac − b2arctan√2ax + b
4ac − b2 + CTrong â ta °t : t = x − p
2ax + b
√4ac − b2L¤i câ :
Z
xdx
ax2+ bx + c =
12a
Z
2ax + b − b
ax2+ bx + cdx =
12a
Z
2ax + b
ax2+ bx + c dx−
b2a
p
2 −√2
−
p
2 −√2
p
2 +√2
−
p
2 +√2
Trang 10⇒ x = cos
π + k2π8
+ i sin
π + k2π8
p
2 +√
2 ,cos5π
8 = cos
7π
8 = −
12
+
+14
+ C, k = 0, , 7
Qua c¡c th½ dö tr¶n ta nhªn th§y c¡c h» sè : α, β, χ, δ , l c¡c h» sè "b§t ành" tùc l "nhúng h»
sè khæng h· thay êi " dò ta câ thay êi "gi¡ trà cõa bi¸n x" º t¼m hiºu rã v§n · n y ta s³
l m quen d¤ng to¡n n y vîi c¡c b i tªp m¨u t÷ìng tü nh÷ sau :
B i 1 : T¼m nguy¶n h m I1 =
Z
x + 27(x − 3) (x + 3)dxGi£ sû r¬ng : x + 27
Z
x + 27(x − 3) (x + 3)dx =
x (α + β) + (4α − 3β)(x − 3) (x + 4)
º tø â ta câ h» :
(
α + β = 84α − 3β = −17 ⇔
(
α = 1
β = 7Vªy :
x2− 6x − 16 =
7x − 26(x − 8) (x + 2) =
(
α + β = 72α − 8β = −26 ⇔
(
α = 3
β = 4Vªy :
Trang 11
α + β + χ = 75β − 5χ = 75
(x − 5)2 =
α
x − 5+
β(x − 5)2 =
α (x − 5) + β(x − 5)2 =
αx − 5α + β(x − 5)2C¥n b¬ng h» sè 2 v¸ ta câ h» :
x3+ 2x2+ x =
6x2+ 20x + 9x(x + 1)2 =
α
x + 1 +
β(x + 1)2 +
χx
Trang 12B i 8 : T¼m nguy¶n h m I8 =
Z
9x3− 20x2+ 30x − 97(x − 2) (x − 3) (x2+ 5)dxGi£ sû r¬ng : 9x3− 20x2+ 30x − 97
( 1
5δ =
35
χx + δ
x2+ 7 =
5(x − 3)2 +
χx + δ
x2+ 7Cho x = 0 ta câ : 17
(x − 3)2(x2+ 7) =
5(x − 3)2 +
Trang 13β(x − a)n−1 + +
χ
x − aTrong â bªc cõa T (x) nhä hìn bªc cõa K (x). º t¼m c¡ch h» sè α, β, , χ ta l m nh÷ sau :
5 ilim
T÷ìng tü ta công câ : lim
12 i
+Im
− 1
12 +
5√5
6 i
Trang 14= 1
10+
√6
√
6 =
15
10x(x2+ 6)2
= − 20
100 = −
15
B i 10 : T¼m nguy¶n h m I10=
Z
x + 1(x2− x + 1) (x2+ 1)dxGi£ sû r¬ng : x + 1
2±
√3
2 i
√32
Z
x + 1(x2− x + 1) (x2+ 1)dx =
Trang 15dx
x2− x + 1 +
12
Z
dx
x − 12
2+34
+ 12
B i 4 : T¼m nguy¶n h m I4 =
Z
x2− 1(x3+ 2) (x − 1)3dxPH×ÌNG PHP L×ÑNG GIC HÂA TRONG TCH PH N
∪
π; 3π 2
Z
sin tdt = cos t + c, π ≤ t < 3π
2 ,
Trang 16(sin t + cos t)2− 1sin t + cos t dt
t +π4
= 2 (sin t − cos t) −√
2 ln
+ c
B i 3 : T½nh t½ch ph¥n A3 =
√ 3 2
2 ⇒ t = π
3Vªy : A3 =
π 3
Z
0
1 − sin2t p1 − sin2t cos tdt =
π 3
Z
0
1 − sin2tcos2tdt =
π 3
Z
0cos4tdt
= 14
π 3
Z
0
1 + 2 cos 2t + cos22tdt
= 14
π 3
π 3
Z
0(3 + 4 cos 2t + cos 4t) dt
= 18
3t + 2 sin 2t + 1
4sin 4t
π 3 0
= π
8 +
7√364C¡ch 2 :
Trang 17Vîi i·u ki»n : 0 ≤ x ≤
√3
2 ta câ :p
1 − x2 =√
1 − x.√
1 + xChó þ r¬ng : 1 − x + 1 + x = 2
2 ⇒ t = π
12Vªy : A3 = −32
π 12
Z
π 4
sin2t.cos2t sin t cos t sin t cos tdt = −32
π 12
Z
π 4
sin4t.cos4tdt
= −2
π 12
Z
π 4
sin42tdt = −2
π 12
Z
π 4
1 − cos 4t2
2
dt = −1
2
π 12
Z
π 4
π 12 π 4
= π
8 +
7√364
2 − x2
°t : x = √2 sin t ⇒ dx =√
2 cos tdt êi cªn : x = 0 ⇒ t = 0, x = 1 ⇒ t = π
4,Vªy : A4 =
π 4
π 4
Z
0
dtcos2t
= 1
2tan t
π 4 0
= 12C¡ch 2 :
Vîi i·u ki»n : 0 ≤ x ≤ 1 ta câ : p
π 8
Z
π 4
−4√2 sin t cos tdt8sin2t.cos2t.2√
2 sin t cos t = −
14
π 8
Z
π 4
dtsin2t.cos2t = −
14
π 8
Z
π 4
1 + 1tan2t
d (tan t)
= −14
tan t − 1
tan t
π 8 π 4
Trang 18Z
− 1 2
√3