MËT SÈ K THUT "CHUÈI" M TA TH×ÍNG GP KHI I TM NGUYN HM HOC TNH TCH PH N
Th½ dö 1 : T¼m nguy¶n h m A1 =
Z
dx (x + 1) (x + 2)
Ta gi£ sû r¬ng : 1
(x + 1) (x + 2) =
α
x + 1 +
β
x + 2
Ta i t¼m 2 h» sè α , β theo hai c¡ch nh÷ sau :
C¡ch 1 : α = lim
x→−1
x + 1 (x + 1) (x + 2) = limx→−1
1
x + 2 =
1
−1 + 2 = 1
β = lim
x→−2
x + 2 (x + 1) (x + 2) = limx→−2
1
x + 1 =
1
−2 + 1 = −1 C¡ch 2 :
Cho x = 0 ta câ : 1
(0 + 1) (0 + 2) =
α
0 + 1 +
β
0 + 2 ⇔ 1
2 = α +
1
2β
x = 1 ta câ : 1
(1 + 1) (1 + 2) =
α
1 + 1 +
β
1 + 2 ⇔ 1
6 =
1
2α +
1
3β
Do â m ta suy ra 2 h» sè α , β b¬ng c¡ch i gi£i h» :
α + 1
2β =
1 2 1
2α +
1
3β =
1 6
⇔
(
α = 1
β = −1 C¡ch 3 :
Ta gi£ sû r¬ng : 1
(x + 1) (x + 2) =
α
x + 1 +
β
x + 2
(x + 1) (x + 2) =
α (x + 2) + β (x + 1) (x + 1) (x + 2) =
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 = 0 2α + β = 1 ⇔
(
α = 1
β = −1
Do â m ta suy ra : 1
(x + 1) (x + 2) =
1
x + 1 − 1
x + 2 Vªy :
Z
dx (x + 1) (x + 2) =
Z
dx
x + 1 −
Z
dx
x + 2 = ln |x + 1| − ln |x + 2| + c = ln
x + 1
x + 2
+ c Th½ dö 2 : T¼m nguy¶n h m A2 =
Z
x + 2
x (x − 2) (x + 5)dx
Ta gi£ sû r¬ng : x + 2
x (x − 2) (x + 5) =
α
x +
β
x − 2 +
χ
x + 5 C¡ch 1 :
α = lim
x→0
x (x + 2)
x (x − 2) (x + 5) = limx→0
x + 2 (x − 2) (x + 5) =
0 + 2 (0 − 2) (0 + 5) = −
1 5
β = lim
x→2
(x − 2) (x + 2)
x (x − 2) (x + 5) = limx→2
x + 2
x (x + 5) =
2 + 2
2 (2 + 5) =
2 7
χ = lim
x→−5
(x + 5) (x + 2)
x (x − 2) (x + 5) = limx→−5
x + 2
x (x − 2) =
−5 + 2
−5 (−5 − 2) = −
3 35 C¡ch 2 :
Cho x = −1 ta câ : −1 + 2
−1 (−1 − 2) (−1 + 5) =
α
−1 +
β
−1 − 2 +
χ
−1 + 5 ⇔ −α −
1
3β +
1
4χ =
1 12
x = 1 ta câ : 1 + 2
1 (1 − 2) (1 + 5) =
α
1 +
β
1 − 2 +
χ
1 + 5 ⇔ α − β +1
6χ = −
1 2
Trang 2x = 3 ta câ : 3 + 2
3 (3 − 2) (3 + 5) =
α
3 +
β
3 − 2 +
χ
3 + 5 ⇔ 1
3α + β +
1
8χ =
5 24
º t¼m ÷ñc c¡c h» sè α , β , χ ta gi£i h» ph÷ìng tr¼nh :
−α − 1
3β +
1
4χ =
1 12
α − β +1
6χ = −
1 2 1
3α + β +
1
8χ =
5 24
⇔
α = −1
5
β = 2 7
χ = − 3 35 C¡ch 3 :
Gi£ sû r¬ng : x + 2
x (x − 2) (x + 5) =
α
x +
β
x − 2 +
χ
x + 5
x (x − 2) (x + 5) =
α (x − 2) (x + 5) + βx (x + 5) + χx (x − 2)
x (x − 2) (x + 5)
x (x − 2) (x + 5) =
x2(α + β + χ) + x (3α + 5β − 2χ) − 10α
x (x − 2) (x + 5)
C¥n b¬ng c¡c h» sè 2 v¸ m ta câ h» :
α + β + χ = 0 3α + 5β − 2χ = 1
−10α = 2
⇔
α = −1 5
β = 2 7
χ = −3 35
Do â : x + 2
x (x − 2) (x + 5) = −
1 5x +
2
7 (x − 2) − 3
35 (x + 5) Vªy :
Z
x + 2
x (x − 2) (x + 5)dx = −
1 5
Z
dx
x +
2 7
Z
dx
x − 2− 3
35
Z
dx
x + 5
= −1
5ln |x| +
2
7ln |x − 2| −
3
35ln |x + 5| + c Th½ dö 3 : T¼m nguy¶n h m A3 =
Z
x2 (−3x2− 2x + 5) (x + 1)dx
Ta c¦n nhî ph÷ìng tr¼nh bªc 2 câ d¤ng : ax2
+ bx + c câ 2 nghi»m x1, , x2 th¼ ÷ñc biºu di¹n d÷îi d¤ng : ax2+ bx + c = a (x − x1) (x − x2)
Do â ta vi¸t : −3x2− 2x + 5 = −3 (x − 1)x +5
3
Ta gi£ sû r¬ng : x2
(−3x2− 2x + 5) (x + 1) = −
x2
3 (x − 1)
x + 5 3
(x + 1)
x − 1+
β
x + 5 3
x + 1
α = lim
x→1
2(x − 1)
3 (x − 1)
x +5 3
(x + 1)
= lim x→1
2 3
x + 5 3
(x + 1)
= − 1
16
β = lim
x→− 5
3
−
x2x +5
3
3 (x − 1)
x + 5 3
(x + 1)
= lim x→− 5 3
2
3 (x − 1) (x + 1)
= −25 48
χ = lim
x→−1
2(x + 1)
3 (x − 1)
x + 5 3
(x + 1)
= lim x→−1
2
3 (x − 1)
x +5 3
= 1 4 C¡ch 2 :
Trang 3Cho x = 0 ta câ : x2
(−3x2− 2x + 5) (x + 1) =
α
x − 1 +
β
x +5 3
x + 1 ⇔ −α + 3
5β + χ = 0
x = 2 ta câ : x2
(−3x2− 2x + 5) (x + 1) =
α
x − 1 +
β
x + 5 3
x + 1 ⇔ − 4
33 = α +
3
11β +
1
3χ
x = 3 ta câ : x2
(−3x2− 2x + 5) (x + 1) =
α
x − 1 +
β
x + 5 3
x + 1 ⇔ − 9
112 =
1
2α +
3
14β +
1
4χ
º tø â m ta câ h» :
−α + 3
5β + χ = 0
α + 3
11β +
1
3χ = −
4 33 1
2α +
3
14β +
1
4χ = −
9 112
⇔
α = − 1 16
β = −25 48
χ = 1 4 C¡ch 3 :
Gi£ sû r¬ng : x2
(−3x2− 2x + 5) (x + 1) = −
x2
3 (x − 1)
x + 5 3
(x + 1)
x − 1 +
β
x +5 3
x + 1
2
3 (x − 1)
x + 5 3
(x + 1)
= α
x +5 3
(x + 1) + β (x − 1) (x + 1) + χ (x − 1)
x +5 3
(x − 1)
x +5 3
(x + 1)
⇔ x2 = x2(α + β + χ) + x8
3α +
2
3χ
+ 5
3α − β −
5
3χ
C¥n b¬ng c¡c h» sè 2 v¸ ta câ h» :
α + β + χ = 1 8
3α +
2
3χ = 0 5
3α − β −
5
3χ = 0
⇔
α = − 1
16
β = −25 48
χ = 1 4
Do â m ta câ :
x2
(−3x2− 2x + 5) (x + 1) = −
x2
3 (x − 1)x +5
3
(x + 1)
16 (x − 1)− 25
48x + 5
3
4 (x + 1) Suy ra :
Z
x2 (−3x2− 2x + 5) (x + 1)dx = −
1 16
Z
dx
x − 1− 25
48
Z
dx
x +5 3
+ 1 4
Z
dx
x + 4
= − 1
16ln |x − 1| −
25
48ln
x +5 3
+ 1
4ln |x + 4| + c Th½ 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
Do â m ta câ : x2+ 4x + 5 = (x + 2 + i) (x + 2 − i)
Ta gi£ sû r¬ng :
x − 1
(x2+ 4x + 5) (x2− 4) =
x − 1 (x + 2 + i) (x + 2 − i) (x2− 4) =
α
x + 2 + i+
β
x + 2 − i+
χ
x − 2 +
δ
x + 2
α = 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) = −
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) = −
13
34+
1
34i
χ = lim
x→2
(x − 1) (x − 2) (x2+ 4x + 5) (x − 2) (x + 2) = limx→2
x − 1 (x2+ 4x + 5) (x + 2) =
1 68
δ = lim
x→−2
(x − 1) (x + 2) (x2+ 4x + 5) (x − 2) (x + 2) = limx→−2
x − 1 (x2+ 4x + 5) (x − 2) =
3 4
Do â m ta câ : x − 1
(x2+ 4x + 5) (x2− 4) =
−13
34− 1
34i
x + 2 + i +
−13
34 +
1
34i
x + 2 − i +
1 68
x − 2+
3 4
x + 2
= −13x − 27
17 (x2+ 4x + 5)+
1
68 (x − 2)+
3
4 (x + 2) C¡ch 2 :
Gi£ sû r¬ng : x − 1
(x2+ 4x + 5) (x2− 4) =
αx + β
x2+ 4x + 5 +
χ
x − 2 +
δ
x + 2 Cho x = 0 ta câ : 1
5β −
1
2χ +
1
2δ =
1 20
x = 1 ta câ : 1
10α +
1
10β − χ +
1
3δ = 0
x = 3 ta câ : 3
26α +
1
26β + χ +
1
5δ =
1 65
x = 4 ta câ : 4
37α +
1
37β +
1
2χ +
1
6δ =
1 148
º tø â m ta câ ÷ñc h» :
1
5β −
1
2χ +
1
2δ =
1 20 1
10α +
1
10β − χ +
1
3δ = 0 3
26α +
1
26β + χ +
1
5δ =
1 65 4
37α +
1
37β +
1
2χ +
1
6δ =
1 148
⇔
δ = −2
5β + χ +
1 10 1
10α −
1
30β −
2
3χ = −
1 30 3
26α −
27
650β +
6
5χ = −
3 650 4
37α −
22
555β +
2
3χ = −
11 1110
⇔
α = −13
17
β = −27 17
χ = 1 68
δ = 3 4
Do â : x − 1
(x2+ 4x + 5) (x2− 4) =
−13x − 27β
17 (x2+ 4x + 5) +
χ
68 (x − 2) +
3δ
4 (x + 2) C¡ch 3 :
Gi£ sû : x − 1
(x2+ 4x + 5) (x2− 4) =
αx + β
x2+ 4x + 5 +
χ
x − 2 +
δ
x + 2
⇔ x − 1 = (αx + β) x2− 4+ χ x2+ 4x + 5(x + 2) + δ x2+ 4x + 5(x − 2)
⇔ x − 1 = x3(α + χ + δ) + x2(β + 6χ + 2δ) + x (−4α + 13χ − 3δ) + (−4β + 10χ − 10δ)
C¥n b¬ng c¡c h» sè 2 v¸ ta câ h» :
α + χ + δ = 0
β + 6χ + 2δ = 0
−4α + 13χ − 3δ = 1 4β − 10χ + 10δ = 1
⇔
δ = −α − χ
β + 6χ + 2 (−α − χ) = 0
−4α + 13χ − 3 (−α − χ) = 1 4β − 10χ + 10 (−α − χ) = 1
⇔
δ = −α − χ
−2α + β + 4χ = 0
−α + 16χ = 1
−10α + 4β − 20χ = 1
⇔
α = −13 17
β = −27 17
χ = 1 68
δ = 3 4
Trang 5Do â : x − 1
(x2+ 4x + 5) (x2− 4) =
−13x − 27
17 (x2+ 4x + 5) +
1
68 (x − 2) +
3
4 (x + 2) Vªy :
Z
x − 1 (x2+ 4x + 5) (x2− 4)dx = −
13 34
Z 2x + 54
13
x2+ 4x + 5dx +
1 68
Z
dx
x − 2 +
3 4
Z
dx
x + 2
= −13
34
Z
(2x + 4)
x2+ 4x + 5dx −
1 17
Z
dx
x2+ 4x + 5 +
1 68
Z
dx
x − 2+
3 4
Z
dx
x + 2
= −13
34
Z d x2+ 4x + 5
x2+ 4x + 5 − 1
17
Z
dx (x + 2)2+ 1 +
1 68
Z
dx
x − 2+
3 4
Z
dx
x + 2
= −13
34ln
x2+ 4x + 5 − 1
17arctan (x + 2) +
1
68ln |x − 2| +
3
4ln |x + 2| + c Th½ dö 5 : T¼m nguy¶n h m A5 =
Z
x
x3+ 1dx
Ta gi£ sû r¬ng : x
x3+ 1 =
x (x + 1) (x2− x + 1) =
α
x + 1 +
β
x − 1
2−
√ 3
2 i
x −1
2 +
√ 3
2 i
α = lim
x→−1
x
x2− x + 1 = −
1 3
β = lim
x→ 1
2 +
√
3
2 i
x (x + 1)
x − 1
2+
√ 3
2 i
= 1
6−
√ 3
6 i
χ = lim
x→12−
√
3
2 i
x (x + 1)
x − 1
2−
√ 3
2 i
= 1
6 +
√ 3
6 i
Do â : x
x3+ 1 = −
1
3 (x + 1) +
1
6−
√ 3
6 i
x − 1
2−
√ 3
2 i +
1
6+
√ 3
6 i
x − 1
2+
√ 3
2 i
3 (x + 1) +
x + 1
3 (x2− x + 1) C¡ch 2 :
Gi£ sû : x
x3+ 1 =
α
x + 1 +
βx + χ
x2− x + 1 Cho x = 0 ta câ : 0 = α + χ
x = 1 ta câ : 1
2 =
1
2α + β + χ
x = 2 ta câ : 2
9 =
1
3α +
2
3β +
1
3χ
Do â m ta câ h» :
α + χ = 0 1
2α + β + χ =
1 2 1
3α +
2
3β +
1
3χ =
2 9
⇔
α = −1
3
β = 1 3
χ = 1 3 Vªy : x
x3+ 1 = −
1
3 (x + 1)+
x + 1
3 (x2− x + 1) C¡ch 3 :
Ta công gi£ sû r¬ng : x
x3+ 1 =
α
x + 1 +
βx + χ
x2− x + 1
⇔ x = α x2− x + 1+ (βx + χ) (x + 1)
Trang 6C¥n b¬ng c¡c h» sè m ta câ h» :
α + β = 0
−α + β + χ = 1
α + χ = 0
⇔
α = −1
3
β = 1 3
χ = 1 3 Vªy :
Z
x
x3+ 1dx = −
1 3
Z
dx
x + 1 +
1 6
Z
2x + 2
x2− x + 1dx
= −1 3
Z
dx
x + 1 +
1 6
Z
2x + 1
x2− x + 1dx +
1 6
Z
dx
x2− x + 1
= −1 3
Z
dx
x + 1 +
1 6
Z d x2− x + 1
x2− x + 1 +
1 6
Z
dx
x −1 2
2 + 3 4
= −1
3ln |x + 1| +
1
6ln
x2− x + 1 + 1
3√
3arctan
2x − 1
√
3 + c Th½ dö 6 : T¼m nguy¶n h m A6 =
Z
dx
x8+ 1 Gi£ sû r¬ng : x8+ 1 = x4+ px2+ 1 x4− px2+ 1= x8+ 2 − p2x4+ 1
çng nh§t 2 v¸ ta ÷ñc : 2 − p2= 0 ⇔ p = ±√
2
⇒ x8+ 1 = x4+√
2x2+ 1 x4−√2x2+ 1
Ta câ : x4+√
2x2+ 1 = x2+ qx + 1 x2− qx + 1= x4+ 2 − q2x2+ 1
çng nh§t 2 v¸ ta câ : 2 − q2 =√
2 ⇔ q = ±p2 −√
2
⇒ x4+√
2x2+ 1 =x2+p2 −√
2x + 1 x2−p2 −√
2x + 1
Ta câ : x4−√2x2+ 1 = x2+ rx + 1 x2− rx + 1= x4+ 2 − r2x2+ 1
çng nh§t 2 v¸ ta câ ÷ñc : 2 − r2
= −√
2 ⇔ r = ±p2 +√
2
⇒ x4−√2x2+ 1 =
x2+p2 +√
2x + 1
x2−p2 +√
2x + 1
Do â m ta câ :
x8+ 1 =
x2+p2 −√
2x + 1
x2−p2 −√
2x + 1
x2+p2 +√
2x + 1
x2−p2 +√
2x + 1
x8+ 1 =
1 8
p
2 −√ 2x + 2
x2+p2 −√
2x + 1
+1 8
−p2 −√
2x + 2
x2−p2 −√
2x + 1
+1 8
p
2 +√ 2x + 2
x2+p2 +√
2x + 1
+
+1
8
−p2 +√
2x + 2
x2−p2 +√
2x + 1
Ta chó þ :
Z
dx
ax2+ bx + c =
1 a
Z
dx
x2+ b
ax +
c a , 4ac − b2> 0
= 1
a
Z
dx
x − −b 2a
2 +
p
4ac − b2 2a
2 = 1 a
Z
dx
°t : p = −b
2a, q =
√ 4ac − b2 2a , x = p + qt
Do â : (1) = 1
aq
Z
1
t2+ 1 dt =
2
√
2
Z
1
t2+ 1 dt =
2
√
2 arctan t
Trang 7= √ 2
4ac − b2arctan√2ax + b
4ac − b2 + C Trong â ta °t : t = x − p
2ax + b
√ 4ac − b2 L¤i câ :
Z
xdx
ax2+ bx + c =
1 2a
Z
2ax + b − b
ax2+ bx + cdx =
1 2a
Z
2ax + b
ax2+ bx + cdx −
b 2a
Z
1
ax2+ bx + c dx
= 1
2aln
ax2+ bx + c − b
2a
Z
dx
ax2+ bx + c + C Vîi : a = c = 1, ∆ = 4 − b2 > 0 ta câ :
Z
Ax + B
x2+ bx + 1dx = A
Z
xdx
x2+ bx + 1 + B
Z
dx
x2+ bx + 1
= 2B − Ab√
4 − b2 arctan√2x + b
4 − b2 + A
2 ln
x2+ bx + 1 + C
Do â : A6.1 = 1
8
2 −√ 2x + 2
x2+p2 −√
2x + 1
dx =
=
p
2 +√ 2
8 arctan
2x +p2 −√
2
p
2 +√ 2 +
p
2 −√ 2
16 ln
x2+p2 −√
2x + 1
+ C1
A6.2= 1
8
Z −p2 −√
2x + 2
x2−p2 −√
2x + 1
dx =
=
p
2 +√ 2
8 arctan
2x −p2 −√
2
p
2 +√ 2
−
p
2 −√ 2
16 ln
x2−p2 −√
2x + 1+ C2
A6.3=
2 +√ 2x + 2
x2+p2 +√
2x + 1
dx =
=
p
2 −√ 2
8 arctan
2x +p2 −√
2
p
2 −√ 2 +
p
2 +√ 2
16 ln
x2+p2 +√
2x + 1
+ C3
A6.4=
Z −p2 +√
2x + 2
x2−p2 +√
2x + 1
dx =
=
p
2 −√ 2
8 arctan
2x −p2 −√
2
p
2 −√ 2
−
p
2 +√ 2
16 ln
x2−p2 +√
2x + 1+ C4
Do â m ta câ : A6 = A6.1+ A6.2+ A6.3+ A6.4
C¡ch 2 :
Ta câ : x8
+ 1 = 0 ⇔ x8 = −1 = cos (π + k2π) + i sin (π + k2π)
⇒ x = cos
π + k2π 8
+ i sin
π + k2π 8
, vîi : k = 0, , 7
Ta câ : sinπ
8 = sin
7π
8 = cos
3π
8 =
1 2
p
2 −√
2, sin3π
8 = sin
5π
8 = cos
π
8 =
1 2
p
2 +√
2 , cos5π
8 = cos
7π
8 = −
1 2
p
2 −√
2 ,
Do â m ta câ :
Z
dx
1 + x8 = −1
8
3
X
k=0
ln
x2− 2x cos
(2k + 1) π 8
+ 1
× cos
(2k + 1) π 8
+
Trang 8+1 4
3
X
k=0
arctan
x sin(2k + 1) π
8
1 − x cos(2k + 1) π
8
× sin
(2k + 1) π 8
+ C, k = 0, , 7
... + 5) + βx (x + 5) + χx (x − 2)x (x − 2) (x + 5)
x (x − 2) (x + 5) =
x2(? ? + β + χ) + x (3 α + 5β − 2χ) − 10α
x (x − 2) (x + 5)
CƠn bơng...
x (x − 2) (x + 5) =
α
x +
β
x − +
χ
x +
x (x − 2) (x + 5) =
α (x − 2) (x + 5) + βx (x +...
x→0
x (x + 2)
x (x − 2) (x + 5) = limx→0
x + (x − 2) (x + 5) =
0 + (0 − 2) (0 + 5) = −
1