Tuyºn tªp c¡c chuy¶n · v k¾ thuªt t½nh nguy¶n hm,t½ch ph¥n. Nguy¹n Duy «ng
MËT SÈ K THUT "PHN TCH " M TA TH×ÍNG GP KHI I TM
NGUYN HM HOC TNH TCH PHN.
Th½ dö 1
: T¼m nguy¶n hm
A1=Zdx
(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 ba c¡ch nh÷ sau :
C¡ch 1 :
α= lim
x→−1
x+ 1
(x+ 1) (x+ 2) = lim
x→−1
1
x+ 2 =1
−1+2 = 1
β= lim
x→−2
x+ 2
(x+ 1) (x+ 2) = lim
x→−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
⇔1
(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 :
Zdx
(x+ 1) (x+ 2) =Zdx
x+ 1 −Zdx
x+ 2 = ln |x+ 1| − ln |x+ 2|+c= ln
x+ 1
x+ 2+c
Th½ dö 2
: T¼m nguy¶n hm
A2=Zx+ 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) = lim
x→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) = lim
x→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) = lim
x→−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
1
Tuyºn tªp c¡c chuy¶n · v k¾ thuªt t½nh nguy¶n hm,t½ch ph¥n. Nguy¹n Duy «ng
x= 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+ 2
x(x−2) (x+ 5) =α(x−2) (x+ 5) + βx (x+ 5) + χx (x−2)
x(x−2) (x+ 5)
⇔x+ 2
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 :
Zx+ 2
x(x−2) (x+ 5)dx =−1
5Zdx
x+2
7Zdx
x−2−3
35 Zdx
x+ 5
=−1
5ln |x|+2
7ln |x−2| − 3
35 ln |x+ 5|+c
Th½ dö 3
: T¼m nguy¶n hm
A3=Zx2
(−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)
=α
x−1+β
x+5
3
+χ
x+ 1
α= lim
x→1
−x2(x−1)
3 (x−1) x+5
3(x+ 1)
= lim
x→1
−x2
3x+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−x2
3 (x−1) (x+ 1)=−25
48
χ= lim
x→−1
−x2(x+ 1)
3 (x−1) x+5
3(x+ 1)
= lim
x→−1
−x2
3 (x−1) x+5
3
=1
4
C¡ch 2 :
2
Tuyºn tªp c¡c chuy¶n · v k¾ thuªt t½nh nguy¶n hm,t½ch ph¥n. Nguy¹n Duy «ng
Cho
x= 0
ta câ :
x2
(−3x2−2x+ 5) (x+ 1) =α
x−1+β
x+5
3
+χ
x+ 1 ⇔ −α+3
5β+χ= 0
x= 2
ta câ :
x2
(−3x2−2x+ 5) (x+ 1) =α
x−1+β
x+5
3
+χ
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
+χ
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)
=α
x−1+β
x+5
3
+χ
x+ 1
⇔ − x2
3 (x−1) x+5
3(x+ 1)
=
αx+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)
=−1
16 (x−1) −25
48 x+5
3+1
4 (x+ 1)
Suy ra :
Zx2
(−3x2−2x+ 5) (x+ 1) dx =−1
16 Zdx
x−1−25
48 Zdx
x+5
3
+1
4Zdx
x+ 4
=−1
16 ln |x−1| − 25
48 ln x+5
3+1
4ln |x+ 4|+c
Th½ dö 4
: T¼m nguy¶n hm
A4=Zx−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) = lim
x→−2−i
x−1
(x+ 2 −i) (x2−4) =−13
34 −1
34i
3
Tuyºn tªp c¡c chuy¶n · v k¾ thuªt t½nh nguy¶n hm,t½ch ph¥n. Nguy¹n Duy «ng
β= lim
x→−2+i
(x−1) (x+ 2 −i)
(x+2+i) (x+ 2 −i) (x2−4) = lim
x→−2+i
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) = lim
x→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) = lim
x→−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+χx2+ 4x+ 5(x+ 2) + δx2+ 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
4
Tuyºn tªp c¡c chuy¶n · v k¾ thuªt t½nh nguy¶n hm,t½ch ph¥n. Nguy¹n Duy «ng
Do â :
x−1
(x2+ 4x+ 5) (x2−4) =−13x−27
17 (x2+ 4x+ 5) +1
68 (x−2) +3
4 (x+ 2)
Vªy :
Zx−1
(x2+ 4x+ 5) (x2−4) dx =−13
34 Z2x+54
13
x2+ 4x+ 5 dx +1
68 Zdx
x−2+3
4Zdx
x+ 2
=−13
34 Z(2x+ 4)
x2+ 4x+ 5 dx −1
17 Zdx
x2+ 4x+ 5 +1
68 Zdx
x−2+3
4Zdx
x+ 2
=−13
34 Zdx2+ 4x+ 5
x2+ 4x+ 5 −1
17 Zdx
(x+ 2)2+ 1 +1
68 Zdx
x−2+3
4Zdx
x+ 2
=−13
34 ln x2+ 4x+ 5−1
17 arctan (x+ 2) + 1
68 ln |x−2|+3
4ln |x+ 2|+c
Th½ dö 5
: T¼m nguy¶n hm
A5=Zx
x3+ 1 dx
Ta gi£ sû r¬ng :
x
x3+ 1 =x
(x+ 1) (x2−x+ 1) =α
x+ 1 +β
x−1
2−√3
2i
+χ
x−1
2+√3
2i
α= lim
x→−1
x
x2−x+ 1 =−1
3
β= lim
x→1
2+√3
2i
x
(x+ 1) x−1
2+√3
2i=1
6−√3
6i
χ= lim
x→1
2−√3
2i
x
(x+ 1) x−1
2−√3
2i=1
6+√3
6i
Do â :
x
x3+ 1 =−1
3 (x+ 1) +
1
6−√3
6i
x−1
2−√3
2i
+
1
6+√3
6i
x−1
2+√3
2i
=−1
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=αx2−x+ 1+ (βx +χ) (x+ 1)
5