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Sotuvchi

Saidmuhammadalixon Ataullayev

Ro'yxatga olish sanasi 29 Noyabr 2023

150 Sotish

Ratsional kasrlar

Sotib olish

O‘ZBEKISTON RESPUBLIKASI OLIY TA’LIM, FAN VA
INNOVATSIYALAR VAZIRLIGI
TOSHKENT AMALIY FANLAR UNIVERSITETI
“MATEMATIKA VA KOMPUTER INJINIRING”
KAFEDRASI
“ DASTURLASH ” fanidan
Mavzu:
______________________________________________________________
______________________________________________________________
______________________________________________________________
Bajardi: ___________________ guruhi talabasi
___________ __________________________
Q abul q ildi: ___________ __________________________
Toshkent – 20__ y.KURS ISHI

TOSHKENT AMALIY FANLAR UNIVERSITETI
“MATEMATIKA VA KOMPUTER INJINIRING”
Kafedrasi
“ DASTURLASH ” fanidan
kurs ishi bo`yicha berilgan
texnik topshiriq
Variant ___
Mavzu : _____________________________________________________
_____________________________________________________________
_____________________________________________________________
1. Kurs ishini topshirish muddati: _______________________________
2. Kurs ishini bajarish uchun boshlang’ich ma’lumotlar: Adabiyotlar, nazariy
ma’lumotlar, grafik muharrirlar. _______________________________
3. Tushuntiruv yozuvi tarkibi: Kirish, mavzu bo’yicha nazariy ma’lumotlar, amaliy
qismda berilgan topshiriq varianti bo`yicha grafik ishlanmalar, xulosa, foydalanilgan
adabiyotlar ro’yxati, ilova. _______________________________
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5. Topshiriq berilgan sana: _____________________________________
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________________ fakulteti, ________________________ guruh talabasi
7. Rahbar: __________________________________________________ «Tasdi q layman»
“ Matematika va komputer injiniring ”
kafedrasi mudiri __________ dots .
_____________ ___ _______ 20 ___ y.

Ekspert guruhi baholari
F.I.O Baho Imzo
Rasmiylashtirish
Himoya
Savollar:
O’rtacha baho

RA T SIONAL KASR LAR
Reja:
1. Kasr-ra t sional funksiya larni sodda kasrlarga ajratish
2. Eng sodda ra t sional kasrlarni integrallash
3. Ba’zi irr at sional ifodalarni integrallash
Tayanch so zlar:ʻ ratsional kasr, ko phadning koeffitsiyentlari, noto g‘ri kasr, to g‘ri ʻ ʻ ʻ
kasr, qoldiq, sodda ratsional kasrlar, noma’lum koeffitsientlar usuli, o
ʻ rniga qo yish ʻ usuli,
sodda ratsional kasrlarni integrallash funksiya, kasr-ratsional, integral, koeffisiyent,
integrallash jadvali, rekurent formula
Har qanday ratsional funksiyani ratsional kasr ko’rinishida, ya’ni ikki
ko’phadning nisbati ko’rinishida tasvirlash mumkin:
Q ( x )
 B
0 x m
 B
1 x m
 1
 ...  B
m
f ( x ) A
0 x n
 A
1 x n
 1
 ...  B
n
Muhokamaning umumiyligini cheklamasdan, bu ko’phadlar umumiy ildizga
ega emas deb faraz qilamiz.
Agar suratinng darajasi mahrajning darajasidan past bo’lsa, kasr to’gri kasr
deb ataladi, aks holda noto’g’ri kasr deb ataladi.
Agar kasr no’to’g’ri bo’lsa, suratni maxrajga (ko’phadni ko’phadga bo’lish qoidasi
boyicha) bo’lib, berilgan kasrni ko’phad bilan biror to’g’ri kasrning yig’indisi
ko’rinishida tasvirlash mumkin:
bu yerda M(x) – ko’phad, – to’g’ri kasr.

1 – misol. Noto’g’ri ratsional kasr berilgan bo’lsin:
.
Suratni maxrajiga (ko’phadlarni bo’lish qoidasi boyicha) bo’lib, shuni hosil
qilamiz:
.
Ko’phadlarni integrallash hech qanday qiyinchilik tug’dirmagani uchun, ratsional
kasrlarni integrallashdagi asosiy qiyinchilik to’g’ri ratsional kasrlarni integrallashdan
iboratdir.
Ma`lumki, P
n ( x)=a
0 x n
+a
1 x n-1
+ a
2 x n-2
+....+ a
n-1 x+a
n funksiya
darajali ko’phad deyiladi. Bunda a
0 , a
1 , a
2 .... a
n - ko’phadning koeffisiyentlari,
n - daraja ko’rsatkichi.
Ikki ko’phadning nisbati kasr-ratsional funksiya yoki ratsional kasr deyiladi:
R ( x ) Q m ( x )
 b x m  b x m1 .... 
b
x 
b
m
0 1 m  1
P ( x )
a0
x n  a x n1 .... 
a n
 1
x 
a nn 1
Agar m<n bo’lsa, u
holda ratsional kasr to’g`ri, agar m  n bo’lsa,u holda
ratsional kasr noto’g’ri kasr bo’ladi.
R(x)- r atsional kasr noto’g’ri bo’lgan hollarda kasrning Q
m (x) suratini P
m (x)
maxrajiga odatdagidek bo’lish yo’li bilan uning butun qismini ajratish kerak:
Shunday qilib, noto’g’ri ratsional kasr bo’lgan holda, undan q(x) butun qismni va
r ( x )
to’g’ri kasrni ajratish mumkin. Demak, noto’g’ri ratsional kasrni integrallash
P ( x )
n
ko’phadni va to’g’ri ratsional kasrni integrallashga keltiriladi.
Misol:
R ( x ) 
2x4  3x3  1
x2  x  2
noto’g’ri ratsional kasrni butun qismini ajrating.
Yechish : R(x) - ratsional kasr noto’g’ri kasr, chunki suratning darajasi maxrajning
darajasidan katta (4>2)

Ko’phadlarni bo’lish qoidasi bo’yicha suratni maxrajga bo’lamiz.
 2 x 4 3x3
 1 x 2
 x  2


 2x3
 4 x 2x2  5 x  9
 2 x 4

I. 
  5 x 3
 4 x 2
 1 
5 x 3
 5 x 2
 10 x


 9 x 2
 10 x  1
J. 
 9 x 2
 9 x  18


19 x  19
Shunday qilib,
R(x)  2x2  5x  9

 19 x  19
ni hosil qilamiz. x2  x 
2
Quyidagi ko’rinishdagi kasrlar eng sodda ratsional kasrlar deyiladi.
F A
 a
A
II ; ( K  2 va butunson )
x 
aK
III Ax  B
( D  0)
x2  px 
q
IV Ax 
B
(S  2 va butun sonlar, hamda D 
0)
x2  px 
qS
Bunda A, B - haqiqiy koeffisiyentlar, a, p, q lar ham haqiqiy sonlar.
Ushbu R ( x )  Q
m
( x )
to’g’ri ratsional kasrni qarab chiqamiz, bu kasrning Pn(x)
P
n ( x )
maxraji (x-a) K
, (x 2
+px+q) S
ko’rinishdagi chiziqli va kvadrat ko’paytuvchilarga
yoyiladi, bunda (x-a) K
ko’rinishdagi ko’paytuvchi K karralikdagi haqiqiy ildizga mos
keladi.
(x 2
+px+q) S
ko’rinishdagi ko’paytuvchi
S karralikdagi kompleks qo’shma
ildizlarga mos keladi.

Pn(x)=a
1 (x-
1 ) K1
(x- 
2 ) K2
....(x- 
t ) Kt
-(x 2
+p
1 x+q
1 ) S1 .
(x 2
+p
2 x+ q
2 ) S2
...(x 2
+px+ q
i ) Si
(1)
Har qanday R ( x )  Q
m
( x )
ratsional kasrni I, II, III, IV turdagi oddiy kasrlarning
P
n ( x )
yig’indisi ko’rinishida ifodalash mumkin. Bunda
a) (1) yoyilmaning (x-
 ) ko’rinishdagi ko’paytuvchisiga I turdagi bitta
A
kasr
x 

mos keladi.
2 (1) yoyilmaning (x-
 ) K
ko’rinishdagi ko’paytuvchisiga I va II turdagi K ta
kasr mos keladi.
A

A  A A
q
 x 
  K
 x    K  1  x    K  2  .....   x   
1 2 3
(4) (1) yoyilmasining (x 2
+px+q) ko’rinishdagi ko’paytuvchisiga III turdagi
kasr mos keladi.
1. (1) yoyilmaning (x 2
+px+q) S
ko’rinishdagi ko’paytuvchisiga III va IV turdagi
S ta kasr mos keladi.
A1x  B1 A2 x  B2 Ai x  Bi
  ..... 
;
x2  px 
qe1
x2  px  qe
1 x2  px 
q
-Misol . R ( x ) 
x  2
;
x3  x
x  2 x  2
x  2
R ( x )
 
xx 2
1 
x3 
x
xx 1 x
1
x 
2

A
 B
 C
; A , B , C  ?
x 3
 x
x x  1 x  1
I va II turdagi oddiy kasrlarni integrallash jadval integrallariga keltiriladi.
I .
 A dx 
A 
d (x  a) 
A ln x 
a x  a x  a
II . 
 Adx
 K  A   x  a   K
d ( x 
a ) 
x  a  C

(x  a)K 1A
A  C   C ;
(1  K )(x  a)K 1
 K  1
III. Turdagi integrallarni ko’rib chiqamiz:

Ax  B P 2
dx , ‡—’Љ—D

 q  0x 2
 px 
q 4
Suratda kasrning maxrajidan olingan hosilani ajratamiz.
(x 2
+px+q) 1
=2x+p
Ax  B
A2x  p
 Ap

B
2 x  p
A  Ap  dx
III .

dx 

2 2


dx

dx   B
  ;
x
2
 px 
q 2
 px  q 2 x 2
 px 
q x
2 px 
q x
 2 
Integrallardan birinchisi ln|x 2
+px+q| ga teng. Ikkinchi integralni hisoblash
uchun maxrajida to’liq kvadratni ajratamiz.

 P 2 P2
x 2
+px+qq  x    q  ;
4
2

P2 P2
Bunda
q   0 , chunki shartga ko’ra D

 q  0
4 4
Demak, ikkinchi integral jadval integraliga keladi. Shunday qilib,

 p 
d  x 
 Ax  B
dx
 A
x2
 px 
q    Ap 
  2 
 A
x2  px 
q ln  B  ln 2 2 2
x  px  q 2  2  
 p 
 q  P 2
 x 
4
2

x
 P
 Ap  1
2
 B    arctg  C
2 2
 2 
q  P q  P
4 4
1-Misol . Integralni hisoblang.
J 
 3 x  8
dx ;
x
2
 4 x  8
Yechish: suratda maxrajining hosilasini ajratamiz.
(x 2+4x+8) 1=2x+4
3x 
8
3 2x  4
3
 4  8
2x  4
J 
 dx 
 2 2 3
 dx 
2 dx
dx

x
2 x 2
 4 x  8 2 x 2 x 2
 4 x  8 4 x  8
 4x 
8 
Birinchi integral ln|x 2
+4x+8| ga teng. Ikkinchi integralning maxrajida to’liq
kvadrat ajratamiz.
(x 2
+4x+8)=(x+2) 2
-4+8=(x+2) 2
+2 2
Natijada quyidagini hosil qilamiz.
J =
3 ln | x 2 + 4x + 8 |
+2 
d(x + 2)
 3 ln | x 2
+ 4x + 8 | +arctg x + 2  C; 2 (x + 2) 2
 22 2
2
Endi IV turdagi integralni hisoblaymiz.

Ax + B
dx ; bunda D = P2  q  0
(x2 + px + q) n4
Bunda ham x 2
+px+q uchxadning hosilasini ajratishdan boshlaymiz.
( Ax  B ) dx
IV .
 
x 2
+ px +
q  n  x 2 + px + q  
 2 x  p
A
(2 x  p )  B
 Ap
(2 x  p ) dx
A 

2 2
dx
  x 2
+ px + q  n 2 x2 + px + q n  
 
B


Ap ) 2


p2

n
2 

p  2
 x +  + q - 
4
 2 


 
Birinchi integralni hisoblasak bo’ladi:

 (2 x 
p ) dx   ( x 2
 px  q )  n
d ( x 2
 px 
q )  1
2 n(1  n)
(x 2  px 
q ) n  1
( x  px  q )
Ikkinchi integralni hisoblaymiz:

p 
belgilashlarni kiritamiz. 0  q
 p2 
a 2
deb olamiz. x    t ; dx  dt
4 2
 
 dt
 1
(t 2  a 2
)  t 2
dt
 1
 dt
 1

t 2dt
;
( t
2
a
2
) n a 2
( t
2 
a 2
) n a
2
(t
2
a
2
) n  1 a
2
( t 2
a
2
)
n
Oxirgi integralga bo’laklab integrallash formulasini qo’llaymiz:
U  t dv
 tdt
; (t 2  a2 )n
du 
dt v

1 
d (t 2 
a2 ) 
1
2
(t 2
 a 2 ) n
2(1  n)
(t
2
a
2
)
n 1
 t 2
dt
 t
 1 
dt
(
t
2  a
2
) n
2(1  n)(t
2 
a 2
) n 2(n  1)
(
t 2 
a
2
)
n 1
Agar J
n 

t 2dt
deb belgilasak, quyidagini hosil qilamiz. (
t
2
a
2
)
n
1) J
n
 t

2n  3  Jn
1 2(n 1)a2  (t 2  a2 )n
1 2(n 1)a2
Bu jarayon quyidagi integralni hosil qilgunimizcha davom etadi.
J1


dt

1arct
g t  C t2

a 2 a a
formula rekurent (qaytuvchan) formula deyiladi.

T a ‘ r i f. Ushbu ko’rinishdagi to’g’ri ratsional kasrlar
A
I.
x  a
A
II.  x  a  k(k –butun musbat son k≥2)
III.
Ax  B
x 2  px 
q
Ax  B
IV.

( x 2  px 
q ) k (mahrajning ildizlari kompleks sonlar, ya’ni p
2
 q  0 )
4
( k -butun musbat son, k ≥2;
mahrajning ildizlar kompleks sonlar); I, II, II va IV tipdagi eng
sodda ratsional kasrlar deb ataladi.
I, II va III tipdagi eng sodda kasrlarni integrallashda katta
qiyinchilikka uchramaymiz, shuning uchun ularning integrallarini hech
qanday izohsiz keltiramiz:
K. 
x
 A
a dx  A ln | x  a |  C
II.

A dx  A ln  (x  a)k dx
 A
(x  a)k 1
 C
 (x  a)k
 k  1

A
 C (1 k )(x  a)k 1
III.

IV tipdagi eng sodda kasrlarni integrallash murakkabroq hisoblashni talab qiladi.
Shu tipdagi integral berilgan bo’lsin:
IV. .
Quyidagicha almashtiramiz:
.
Birinchi integral , almashtirish yordami bilan
olinadi:
.
Ikkinchi integralni (uni I
k bilan berlgilaymiz) ushbu ko’rinishda yozamiz:
,
bu yerda
deb faraz qilamiz (maxrajning ildizlari farazga ko’ra kompleks sonlar,
demak ). Endi quyidagicha kirishamiz:
. (1)
So’nggi integralni bunday almashtiramiz:
.

Bo’laklab integrallab, quyidagini hosil qilamiz:
.
Bu ifodani (1) tenglikka qoysak, quyidagi hosil bo’ladi:
.
O’ng tomonda I
k tipidagi integral bor, lekin integral ostidagi funksiya maxrajining
daraja ko’rsatkichi uning daraja ko’rsatkichidan bitta birlik past (k–1), shunday
qilib, I
k ni I
k-1 orqali ifodaladik.
Shu yo’l bilan davom etib, ma’lum integral
ga yetib boramiz. So’gra t va m o’rniga ularning qiymatlarini qo’yib, IV
integralning x va berilgan A,B,p,q sonlar orqali ifodasini topamiz.
2 – misol.
.
Oxirgi integralga x+1=t almashtirishni qo’llaymiz:
.

Oxirgi integralni qaraymiz
Demak,
.
oxirgi natija quyidagi ko’rinishda bo’ladi:
.
Endi har qanday to’gri ratsional kasrni eng sodda kasrlar yig’indisiga
ajratish mumkinligini ko’rsatamiz.
Ushbu to’g’ri ratsional kasr
G
( x ) f
( x )
berilgan bo’lsin. Bu kasrga kirgan ko’phadlarning koeffisentlari haqiqiy sonlar
va berilgan kasr qisqarmaydigan kasr deb faraz qilamiz (bu esa surat va mahraj
umumiy ildizga ega emas degan so’zdir).
1 – t e o r e m a. x=a mahrajning k karrali ildizi, ya’ni f(x)=(x-a) k
f
1 (x) bo’lsin, bu
yerda f
1 (a)≠0; u holda …. To’’ri kasrni boshqa ikki mo’g’ri kasr yig’indisi
shaklida quyidagicha tasvirlash mumkin
F ( x )
 A
F1 (x)
f ( x ) ( x  a ) k
( x  a ) k
 1
f ( x )
1
bu yerda A nolga teng bo’lmagan o’zgarmas son, F
1 (x) ko’phad, buning
darajasi (x-a) k-1
f
1 (x) mahrajning darajasidan past.

3 – t e o r e m a . Agar f(x)=(x 2
+px+q) φ
1 (x) bo’lsa (bu yerda φ
1 (x)
ko’phad
x 2
+px+q ga bo’linmaydi), …. to’g’ri ratsional kasrni boshqa ikki too’g’ri kasrning
yig’indisi ko’rinishida tasvirlash mumkin:F (x)
 Mx  N
 
1 ( x )
f ( x )
(
x
2 px 
q) k
( x
2 px 
q)
1
 ( x )
1
Isbot. Ushbu ayniyatni yozamiz:
(4)
bu ayniyat har qanday M va N uchun to’g’ri; M va N ning shunday qiymatini
topamizki, unda ko’phad x 2
+px+q ga bo’linsin.
Buning uchun
x 2
+px+q ning ildizlariga teng ildizlarga ega bo’lishi zarur va
yetarlidir. Demak,
yoki
lekin aniq bir kompleks son, buni K +iL ko’rinishda yozish mumkin,
bu yerda K va L haqiqiy sonlar. Shunday qilib,
,
bundan

yoki
.
M va N ning shu qiymatlarida ko’phad ildizga,
demak, qo’shma ildizga ham ega bo’ladi. Bu holda ko’phad
va ayirmalarga, demak, ularning ko’paytmasiga, ya’ni x 2
+px+q ga
qoldiqsiz bo’linadi. Bo’linmani Ф
1 (x) bilan belgilab, quyidagicha yoza olamiz:
.
(5) tenglikdagi ohirgi kasrni x 2
+px+q ga qisqartirib, (3) tenglikni hosil
qilamiz, bunda Ф
1 (x) ning darajasi maxrajning darajasidan past ekani ravshan.
Shuni isbot qilish talab qilingan edi.
1. Kasr-ra t sional funksiya lar ni sodda kasrlarga ajratish
Ma’lumki, Pn(x)=a0xn+a1xn−1+a2xn−2+...+an−1x+an
funksiya n- darajali ko phad deyiladi, bunda
ʻ a0,a1,a2,...,an –ko phadning ʻ
koeffitsiyentlari.
Quyidagi
n n n n m m m m
n
m
b x b xa x a
b x b xb x b
x P
x Q x R    
     
   
1 1 1 0
1 1 1 0
...
...
) (
) ( ) (
(11.1)
ko rinishdagi kasrga
ʻ kasr-ratsional funksiya yoki qisqacha ratsional kasr deyiladi.
Bu yerda
N n m  , va m j n i R b a j i ,1 , ,1 , ,    .
Agar
n m  bo lsa, ʻ to g‘ri kasr ʻ , n m  bo lsa, ʻ noto g‘ri kasr ʻ deyiladi.
Har qanday noto g‘ri kasr ko phadlarni bo lish qoidasi yordamida qandaydir
ʻ ʻ ʻ
ko phad va to g‘ri kasr yig‘indisi shaklida ifodalanadi:
ʻ ʻ
) (
) ( ) ( ) (
) ( ) ( x P
x r x M x P
x Q x R
n nm n
m    
. (11.2)
Masalan,
10 3
12 33 10 3
1 3
2
2
2
2
4
 
     
 

x x
x x x
x x
x , chunki
x 4
+ 2 x 2
+ 3 x – 1

x 4
+ 3 x 3
– x 2
x 2
−3 x +10
−3 x 3
+ x 2
+2
− 3 x 3
–9 x 2
+ 3 x
10 x 2
– 3 x + 2
10 x 2
+ 30 x – 10
– 33 x+ 12
Quyidagi kasrlar eng sodda ra t sional kasrlar deyiladi:
I. A
x−α , II.
A
(x− α)k (k≥ 2 va butun son )
III.
Ax +B
x2+ px +q
(maxrajning diskriminanti D <0) .
І V.
Ax +B
(x2+ px +q)s (s≥ 2 va butun , D <0).
(Bu yerda A, B – biror haqiqiy koeffitsientlar α , p , q lar ham haqiqiy sonlar)
Ma’lumki, har qanday haqiqiy koeffitsientli ko phad quyidagi ko paytma
ʻ ʻ
shaklida ifodalanadi:
sts s t k k n q x p x q xp x x x a x P ) (... ) ( ) (... ) ( ) ( 2 1 1 2 1 0 1 1              
, (11.3)
bu yerda
  ,...,1 lar ko phadning ʻ k k ,...,1 karrali haqiqiy ildizlari,
) ,1 (,0 4 2 s i q p i i   
va n t t k k s      2 ... 2 ... 1 1  .
Teorema ( to g‘ri kasrni sodda kasrlar yig‘ndisiga ajratish
ʻ
haqida) Maxraji (11.1) shaklda tasvirlangan har qanday to g‘ri ratsional kasrni I-
ʻ
IV turdagi oddiy kasrlar yig‘indisiga yoyish mumkin. Bu yoyilmada
) (x Pn
ko phadning har bir
ʻ rk karrali r haqiqiy ildiziga ( ( ) rkr x  ko paytuvchisiga) ʻ
     
3 1 2 2 3
... r
r
k
k r r r r A
A
A A
x
x x x      

  
(11.4)
ko rinishdagi
ʻ rk ta oddiy kasrlar yig‘indisi mos keladi. ) (x Pn ko ʻ phadning har bir
juft qo ʻ shma - kompleks ildiziga (
2 ( ) t x p x q      ko ʻ paytuvchisiga )
     
3 3 1 1 2 2 2 3 2 2 2 2
...
t t
t M x N
M x N
M x N M x N
x p x q
x p x q x p x q x p x q  
         

 
   
 
     
(11.5)
ko rinishdagi
ʻ t ta oddiy kasrlar yig‘indisi mos keladi.
Yoyilmadagi
, ,A M N koeffitsientlarning qiymatlari esa
1) noma’lum koeffitsientlar usuli ;
2) o
ʻ rniga qo yish ʻ usuli dan biri yoki ikkalasini qo llab aniqlanadi. ʻ

Noma’lum koeffitsientlari usulida ( )R x to g‘ri ratsional kasrning suratidagi ʻ
ko phad hosil bo lgan kasrning suratidagi ko phadga aynan tengligidan
ʻ ʻ ʻ x ning bir
xil daragalari oldidagi koeffitsientlar tenglab,
n ta noma’lum uchun n ta
tenglamalar sistemasi hosil qilinib noma’lum koeffitsientlar topiladi.
O rniga qo yish usulida ko phadlar,
ʻ ʻ ʻ x ning barcha qiymatlarida aynan teng
bo lgani uchun,
ʻ x ning tayin xususiy qiymatlarida tenglab noma’lum koeffitsientlar
topiladi.
1-m isol. Ushbu
R(x)= x+2
x3− x r a t sional kasrni oddiy kasrlar yig‘indisiga ajrating.
Ye chish.
R(x) r a t sional kasr to g‘ri kasr, chunki suratning darajasi ʻ
maхrajning darajasidan kichi k . Kasrning maхrajini ko paytuvchilarga ajratamiz:
ʻ
x3− x= x(x+1)(x−1)
Keltirilgan teoremaga asosan,
R(x) kasrni sodda kasrlarga ajratish bunday
ko rinishda bo lishi kerak:
ʻ ʻ
R(x)= x+2
x(x+1)(x−1)
= A
x+ B
x+1+ C
x−1
A , B , C koeffi t si y entlarni topamiz. Buning uchun tenglikning o ng qismini umumiy
ʻ
maхrajga keltiramiz va hosil qilingan tenglikning ikkala qismida maхrajni tashlab
yuboramiz. Bu amallar natijasi quyidagi tenglikdan iborat bo ladi:
ʻ
x+2≡ A(x+1)(x−1)+Bx (x−1)+Cx (x+1)
.
x o zgaruvchiga istalgan uchta haqiqiy sonli qiymat berib,
ʻ A , B , C larga
nisbatan uchta noma’lum uchta tenglama sistemasini hosil qilamiz. Bu sistemani
yechib, noma’lum A , B , C koeffisientlarni topamiz. Sonli qiymatlarni o rniga
ʻ
qo yish usuli ana shundan iborat. Agar
ʻ x o zgaruvchiga maхrajning ildizlari ʻ
qiymati ketma-ket berilsa, yanada sodda tenglamalarni hosil qilamiz, chunki ularda
har gal faqat bitta noma’lum qoladi.
Haqiqatan ham, qat’iy tenglikdagi o zgaruvchiga dastlabki kasr maхrajining
ʻ
ildizlari
0,−1,1 qiymatlarni beramiz. Agar x=0 bo lsa, ʻ 2= A(−1) ni topamiz,
bundan
A=−2 .
Agar x = − 1
bo lsa, ʻ 1= B(−1)(−1−1) ni topamiz, bundan B = 1
2 .
Agar
x=1 bo lsa, ʻ 3 = C ∙ 1 ( 1 + 1 )
bo ladi, bundan ʻ C = 3
2 .

Endi berilgan tenglikni quyidagi ko rinishda yozish mumkin:
ʻ

R(x)= x+2
x(x+1)(x−1)=− 1
x + 1
2(x+1)+ 3
2(x−1)2-m isol. Ushbu
x
x3−1 ra t sional kasrni sodda kasrlar yig‘indisiga ajrating.
Ye chish. Berilgan kasrni sodda kasrlar yig‘indisiga ajratamiz:
x
(x− 1)(x2+ x+1)
= A
x− 1 + Bx +C
x2+ x+ 1
x≡ Ax 2+ Ax + A+ Bx 2+Cx − Bx − C
x≡ x2(A+B)+x(A+C− B)+(A− C )
.
K asrning maхraji faqat bitta haqiqiy ildizga ega (
x=1 ) . Shuning uchun o rniga ʻ
qo yish
ʻ va noma’lum koeffitsientlar usullaridan birdan foydalanib, quyidagi
tenglamalar sistemasini hosil qilamiz va undan esa A , B , C koeffitsientlarni
topamiz:
x=1:1=3A¿}x
2
:0=A+B¿}¿¿¿
,
x
x3−1
= x
(x−1)(x2+x+1)
= 1
3(x−1)− x−1
3(x2+x+1)
2. Eng sodda ra t sional kasrlarni integrallash
Demak, integral ostidagi har qanday
( )R x to g‘ri ratsional kasrni (11.4) va ʻ
(11.5) formulalarni e’tiborga olib noma’lum koeffitsientli oddiy kasrlarga yoyiladi.
So ng bu kasrlardagi noma’lum koeffitsientlar topiladi. So ngra bu sodda kasrlar
ʻ ʻ
integrallanar ekan.
Endi biz bu to rtta eng sodda
ʻ kasrlarning integrallarini hisoblaymiz.
1.
C x A dxa x
A      2 ln .
2.
 
 
  
C
a x k
A dx a x A dx
a x
A kk
k 
 
   
   
1 1
.
3.
dxq px x
B Ax
  

2 integralda 0  A bo lsa, suratida maxrajining hosilasini ʻ
hosil qilib olamiz:
 
2 2 2
2 2 2
2 2
B x p p Ax B A A x p A dx dx dx x px q x px q x px q
                      
   





   
 
 

 
        
 
  
4 2
2 ln2 2 2 2 2 2 p q p x
dx Ap B q px x A
q px x
dx Ap B
.
Oxirgi integralda
 0 0 4
4
4
2 2
     D p q p q bo lgani uchun, jadvaldagi
ʻ   2 2 a u
du
integralga keladi. Demak,

  C
p q
p x arctg
p q
Ap B q px x A dx
q px x
B Ax 



    
 

 2 2
2
2 4
2
4
2 ln
2. (11.6)
4.
      





   
 
 

 
  
 
 
 

k k k
p q p x
dx Ap B dx
q px x
p x A dx
q px x
B Ax
4
4
2
2
2
2 2 2 2 2 .
Bund а
    1
22 )1 (
1
2
2
2    
  
 

kk q px x k
A dx
q px x
p x A
, (11.7)
oxirgi integralda esa
2
4 ,2
2p q a p x u     almashtirish bajaramiz.
 
 
     
  


 

  du
au u
a
au du
adu
au uau
a
au du
kkkk
22 2
21
222
22 222
2
22 111
.
Birinchi integral berilgan integralning tartibi bittaga kamaygan holi, ikkinchi
integralni bo laklab integrallash mumkin. Natijada, quyidagi rekkurent formulani
ʻ
hosil qilamiz:
       




 1
2221
22222
)1(2 32
)1(2 kkk
au du
ka k
auka u
au du
. (11.7)
Eslatma. Agar maxrajda
c bx ax  2 ko phad bo lsa, avval a qavsdan
ʻ ʻ
chiqariladi:


 

     
a
c x
a
b x a c bx ax 2 2
3-misol. Ushbu
dx x x
x
  

26 8 2
2 3
2 integralni hisoblang.
Yechish. Avval maxrajidan
2 ko paytuvchini qavsdan chiqaramiz, suratida ʻ
maxrajining hosilasini hosil qilib integrallaymiz.
    
 

 
 
 
  

 

2 2 2 2 2 3 )2 (
4
13 4
4 2
4
3
13 4
3
4 4 4 2
4
3
13 4
2 3
2
1
x
dx dx
x x
x dx
x x
x
dx
x x
x
  C x arctg x x       3
2
3
4 13 4 ln4
3 2
.◄
4-misol. Ushbu
 
dx
x x
x
  

2 2 5 2
3 7 integralni hisoblang.
Yechish.
             

 
 
 
  

 

22 2 2 2 2 2 2 2 2 )1 (
4
5 2
2 2
2
7
5 2
7
6 2 2 2
2
7
5 2
3 7
x
dx dx
x x
x dx
x x
x
dx
x x
x
.
Birinchi qo shiluvchi (11.6) formulaga ko ra,
ʻ ʻ

 521
27
52 22
27
22
2

  
xxdx
xx x
.
Ikkinchi integral uchun (11.7) rekkurent formulani qo llasak,
ʻ
   

 
 

 

   2222
2222
22
2)1( )1(
)12(22 322
2)1()12(22 1
2)1( x xd
xx
x dx
  2
1
2
1
8
1
5 2 (8
1
2 2
  
 
  x arctg
x x
x
.
Demak,
   
Cx
arctg
xx x
xxdx
xx x


 


  
2 1
16 1
)52(8 1
522 7
52 37
2222
2
.
5-mis o l. Ushbu
2
2
3 2
( 1)
x x dx x x
 
  integralni hisoblang.
Yechish. Maxrajdagi ko phadning
ʻ 0 x bir karrali haqiqiy va 1 x ikki
karrali ildizlari bor bo lgani uchun
ʻ
2
2 2
3 2
( 1) ( 1) 1
x x A B C
x x x x x
       
.
mumiy maxraj berib, suratdagi ko phadlarni tenglaymiz.
ʻ
2 2 2 3 2 2 x x Ax xA A Bx Cx Cx        
yoki
    2 2 3 2 2 x x x A C x A B C A        
.
Noma’lum koeffitsientlari usulidan foydalanamiz,
x ning darajalari oldidagi
koeffitsintlarni tenglaymiz:
. 2 :
;3 2 :
;1 :
0
2

  
 
А x
C B A x
С A x
Bundan ,
1 ,6 ,2    C B А .
Demak,
      
 
    dx x dx x dxx dx x x
x x
1
1
)1 (
6 2
)1 (
2 3
2 2
2
C
xx x
Cx
xx 



16
1ln1ln
16
ln2 2
.
6-mis o l. Ushbu
  

)2 )(1 (
)3 ( 2
x x x
dx x integralni hisoblang.
Yechish. Integral ostida to g‘ri ratsional kasr va u І turdagi sodda kasrlar
ʻ
yig‘indisiga ajraladi
21)2)(1( 32


 
x D
x B
xA
xxx x
,

bundan ).1 ( )2 ( )2 )(1 ( 3 2         x Dx x Bx x x A x
A , B , D koeffutsientlarni topish uchun o rniga qo yish usulidan
ʻ ʻ
foydalanamiz:
0 x
bo lganda ʻ A2 3  , bundan ;2
3 A
1x
bo lganda ʻ , 3 4 B  bundan ;3
4  B
2x
bo lganda, ʻ ,67 D
bundan
.6
7  D
Shunday qilib , quyidagini hosil qilamiz :

 
  
    

2
)2 (
6
7
1
)1 (
3
4
2
3
)2 )(1 (
)3 ( 2
x
x d
x
x d
x
dx
x x x
dx x
C
x
x x C x x x           6 9
7 8 2 )1 ( ln 2 ln6
7 1 ln3
4 ln2
3
.
7-mis o l. Ushbu
 8 3x
dx integralni hisoblang.
Yechish. Integral ostida to g‘ri ratsional kasrning maxrajidagi ko phad
ʻ ʻ
ko paytuvchilarga ajratiladi va sodda kasrlar yig‘indisi shaklida ifodalanadi
ʻ
.4 2 2 )4 2 )(2 (
1
8
1
2 2 3  
         x x
N Mx
x
A
x x x x
Umumiy maxraj berib suratlari tenglanadi
1 )2 ( )2 ( )4 2 ( 2        x C x Bx x x A
A , M , N koeffitsientlarni topish uchun yuqoridagi usullarni birga qo llaymiz:
ʻ
. 124: ;0: 112:2
0 2
  
CАx BAx Ax
Bundan ,
31 , 121 , 121    C B А va
)4 2 ( 12
4
)2 ( 12
1
8
1
2 3  
     x x
x
x x
.
Endi integralni hisoblaymiz:
    
 
     
         dx x x
x x dx x x
x
x
dx
x
dx
4 2
6 2 2
2 12
1 2 ln 12
1
4 2
4
12
1
2 12
1
8
223
 

 
   
      2 2 2 3 )1 (
)1 (
4
1
4 2
)2 2(
24
1 2 ln 12
1
x
x d
x x
dx x x
         C x arctg x x x
3
1
3 4
1 4 2 ln 24
1 2 ln 12
1 2
  C x arctg x x
x     
  3
1
3 4
1
4 2
2 ln 24 2
2
.

Shunday qilib ratsional kasrni integrallash uchun
1) uning to g‘ri yoki noto g‘ri kasr ekanligini tekshiriladi, aks holda(ya’niʻ ʻ
noto g‘ri kasr bo lganda) butun qismi ajratiladi, ko phad va to g‘ri ratsional kasr
ʻ ʻ ʻ ʻ
hosil qilinadi;
2) to g‘ri ratsional kasrni oddiy kasrlar yig‘indisiga ajratiladi;
ʻ
3) yoyilmaning koeffitsientlari topiladi;
4) ifoda integrallanadi.
3. Ba’zi irr at sional ifodalarni integrallash
Har qanday irratsional funksiyalar ucnun ham elementar funksiyalar
ko rinishidagi boshlang‘ich funksiyalarni aniqlab bo lmaydi. Biz quyida ayrim
ʻ ʻ
almashtirishlar yordamida ratsional funksiyalar integrallariga olib kelinadigan
irratsional funksiyalarning integrallarini ko rib chiqamiz.
ʻ .
Quyidagi
dx d cx
b ax
d cx
b ax
d cx
b ax x R nnsr sr sr
 








 


 
 


 
 


 ..., , , 22 11 , (11.8)
bu yerda R -ratsional funksiya, a , b , c , d - o zgarmas sonlar,
ʻ r
i , s
i musbat butun
sonlar, integral m
t
d cx
b ax 


(11.9) almashtirish yordamida ratsionallashtiriladi.
Bu yerda m - nn
s r
s r
s r
...,,
22
11
kasrlarning umumiy maxraji, ya’ni
) ,... , (
21 n s s s EKUB m 
.
Xususan,
dx x x x x R
nn
s r
s r
s r 




 ..., , ,
22
11
Integral m
t x
almashtirish yordamida ratsionallashtiriladi.
8-misol. Ushbu
dx
x
x
  4 3 4 integralni hisoblang.
Yechish.
(2, 4) 4 EKUK  bo lgani uchun, ʻ
     

 



 





    C t t dt
t
t t dt
t
t
dtt dx
t x
dx
x
x 4 ln
3
16
3
4
4
4 4
4
4
4 4
3 3
3
2 2
3
5
3
4
4 3
C x x x t       4 ln
3
16
3
4
4 3
4 3
4
.◄
Integral

dx
c bx ax
B Ax
  

2ko rinishida berilgan bo lsin. Avval kasr suratida ildiz ostidagi kvadrat uchhadning
ʻ ʻ
differensiali hosil qilinadi( 0A
), kvadrat uchhaddan to la kvadrat ajratiladi va
ʻ
quyidagi amallar bajariladi:
  
  
 
  
 
 
 

   c bx ax
dx
a
Ab B
c bx ax
dx b ax
a
A dx
c bx ax
B Ax
2 2 2 2
2
2



 

   

 

 


 

     
a
b c
a
b x a
dx
a
Ab B c bx ax
a
A
4 2
2 2 2
2
.
Agar
0 ,
4 2
  a
a
b с
bo lsa, oxirgi integralni
ʻ
C k u u
k u
du    
  2
2 ln
integralga keltirib hisoblash mumkin.
Agar
0 ,
4 2
  a
a
b с
bo lsa,
ʻ
C
k
u
u k
du  
  arcsin 2 2 integralga keltirib
hisoblash mumkin.
Eslatma. Qulaylik uchun, kvadrat uchhadni to la kvadratga ajratishdan
ʻ
avval
a ning modulini ildizdan chiqarish kerak.
9-misol. Ushbu
dx
x x
x
  

8 4
3 5
2 integralni hisoblang.
Yechish.

  
 
   
 
 
 

   8 4 2
4 5 3
8 4
4 2
2
5
8 4
3 5
2 2 2 x x
dx dx
x x
x dx
x x
x
 
  C x x x x
x
dx x x         
 
     4 2 2 ln7 8 4 5
4 2
7 8 4 5 2 2
2
2
.
10-misol. Ushbu
dx
x x
x
  

2 2 8 10
2 3 integralni hisoblang.
Yechish. Qulaylik uchun, avval 2 ni ildizdan chiqarib olamiz.
1 2 2 2
2
4 5
2 3
2
1
2 8 5
21 3 I dx
x x
x dx
x x
x
  
 
 
 

.
Hosil bo lgan integralni hisoblaymiz.
ʻ

    
 
   
 
   

 
     2 2 2 1 4 5
2 4
2
3
4 5
8 2 4
2
3
4 5
2 3
x x
dx x dx
x x
x
dx
x x
x I
 .)2arcsin(8453
218
12
2 Cxxx
xdx



Demak,
C x x x dx
x x
x       
 

 )2 arcsin(2 4 4 5
2
2 3
2 8 10
2 3 2
2
.
Agar integral
dx
c bx ax x
B Ax
   

2 ) ( 
ko rinishda bo lsa,
ʻ ʻ tx 1


almashtirish yordamida hisoblanadi.
11-misol. Ushbu
    10 2 )1 ( 2 x x x
dx integralni hisoblang.
Yechish.

  




 
    1 9 9 1 1
1
1
1 1
10 2 )1 ( 2
2
2
2
2 t
dt
t t
dt t
dt t dx
t x
x x x
dx
 
C
x x
C t t  



       1
1
9
1
3 ln
3
1 1 9 3 ln
3
1
2 2
.
Agar integral
 
dxcbxaxxR
  2,
ko rinishda bo lsa, kvadrat uchhadni to la kvadratga ajratib, quyidagi
ʻ ʻ ʻ
1)
 du u k u R  22 ,
, )cos(sin tkutku 
almashtirish;
2)
 du u k u R  22 ,
, ) ( kctgt u ktgt u   almashtirish;
3)
 du k u u R  22 ,
, 

 

  
t
k u
t
k u
cos sin almashtirish
yordamida hisoblanadigan integrallardan biriga keltirish mumkin.
12-misol. Ushbu
   dx x x 2 2 3 integralni hisoblang.
Yechish.


 
        tdt dx
t x
dx x dx x x
cos2
sin2 1
)1 ( 4 2 3 2 2

             C t t dtt tdt dtt t 2 sin 2 ) 2 cos 1( 2 cos 4 cos2 sin4 422
C x x x x C t t t           
2
2 3 )1 (
2
1 arcsin2 sin 1 sin2 2
2 2
.
13-misol. Ushbu
 

  3
2 5 4x x
dx
integralni hisoblang.
Yechish.

       




 

 

 
   3 2 2 2
3 2 3 2 1 cos cos
2
1 2 5 4 t tg t
dt
t
dt dx
tgt x
x
dx
x x
dx
 
C
x x
x C
t tg
tgt C t tdt
t tg t
dt 
 
  

   

   5 4
2
1
sin cos
1 cos 2 2 3 2 2
.
Ratsional kasrlar va ularni integrallash
Ta’rif. Ikkita ko’phad nisbatidan iborat funksiya ratsional kasr yoki
ratsional funksiya deyiladi. Odatda u R ( x )
bilan belgilanadi.
R(x)= Qm(x)
Pn(x)= bmxm+bm−1xm−1+… +b1x+b0
anxn+an−1xn−1+… +a1x+a0
Ta’rif. Agar R ( x )
ratsional kasrda maxrajining darajasi
n>m bo’lsa, u
to’g’ri, n ≤ m
bo’lsa, noto’g’ri ratsional kasr deb ataladi.
Agar
R(x) noto’g’ri ratsional kasr bo’lsa, u holda uni quyidagicha
yozish mumkin:
R
( x ) = Q
m ( x )
P
n ( x ) = N
m − n ( x ) + G
r ( x )
P
n ( x ) , r < n .
Ta’rif. R
1
( x ) = A
x − a , II. R2(x)= A
(x− a)k , III. R
3 ( x ) = Ax + B
x 2
+ px + q ,
IV.
R4(x)= Ax + B
(x2+ px +q)k ko’rinishdagi kasrlar eng sodda ratsional
kasrlar deb ataladi. Bu yerda
A,B,a,p,q−¿ haqiqiy sonlar, k=2,3,4 ,… va
x2+px +q
kvadrat uchhad haqiqiy ildizlarga ega emas, ya’ni D < 0
deb
olinadi.
I.
 R1(x)dx = A
x− adx = Aln |x− a|+c.

II.  R2(x)dx = A
(x− a)kdx = A (x− a)−kd(x−a)=¿¿
¿ A
(1− k)(x−a)k−1+c.

III.
 R
3
( x ) dx =
 Ax + B
x 2
+ px + q dx = ¿
 A
2
( 2 x + p ) + ( B − Ap
2 )
x 2
+ px + q dx = ¿ ¿ =
A
2 (2x+p)dx
x2+px +q +(B− Ap
2 ) dx
x2+px +q=¿{
x2+px +q=t
(2x+p)dx =dt }¿
⟹
A
2  dt
2 + + ( B − Ap
2 )
 dx
x 2
+ px + q = ¿ A
2 ln
| x 2
+ px + q | + ( B − Ap
2 )  d
( x + p
2 )
(
x + p
2 ) 2
+ m 2 = A
2  dt
2 + + ( B − Ap
2 )
 dx
x 2
+ px + q = ¿ A
2 ln
| x 2
+ px + q | + ( B − Ap
2 )  d
( x + p
2 )
(
x + p
2 ) 2
+ m 2 = A
2  dt
2 + + ( B − Ap
2 )
 dx
x 2
+ px + q = ¿ A
2 ln
| x 2
+ px + q | + ( B − Ap
2 )  d
( x + p
2 )
(
x + p
2 ) 2
+ m 2 = ¿ ¿ ¿ ¿
¿A
2ln|x2+px +q|+(B− Ap
2 )
1
m arctg
x+ p
2
m +C .
IV.
 R4(x)dx = Ax +B
(x2+px +q)kdx =
A
2(2x+p)+(B− Ap
2 )
(x2+px +q)k dx =¿¿
¿A
2 (2x+p)dx
(x2+px +q)k+(B− Ap
2 )
d(x+ p
2)
[(x+ p
2)
2
+m2]
kdx =¿¿
¿ A
2 
( x 2
+ px + q ) − k
d ( x 2
+ px + q ) + ¿
+(B− Ap
2 )
d(x+ p
2)
[(x+ p
2)
2
+m2
]
k=¿ 1
(1− k)(x2+px +q)k−1+¿¿
+
( B − Ap
2 )  dt (
t 2
+ m 2 ) k = 1 (
1 − k )( x 2
+ px + q ) k − 1 + ¿
+(B− Ap
2 )
m2  t2+m2−t2
(t2+m2)kdt = 1
(1− k)(x2+px +q)k−1+¿
+(B− Ap
2 )
m2  dt
(t2+m2)k−1−
(B− Ap
2 )
m2  t2dt
(t2+m2)k.
Bu yerda ikkinchi qo’shiluvchi maxrajida k − 1
darajali kvadrat
uchhad qatnashgan integraldan iborat. Oxirgi qo’shiluvchini bo’laklab
integrallash formulasini qo’llab, undan ham maxrajida
k−1 darajali
kvadrat uchhad qatnashgan integral hosil qilinadi. Hosil bo’lgan

integralda yana yuqoridagi ishlar takrorlanadi va pirovard natijadaR1(x),R2(x)
va R3(x) ko’rinishdagi integrallarga keltiriladi.
Mavzuga doir yechimlari bilan berilgan topshiriqlardan namunalar
1.  5 dx
x + 4 integral hisoblansin.
Yechish:  5 dx
x + 4 = 5
 d ( x + 4 )
x + 4 = 5 ln
| x + 4 | + C
.
2.  13 dx
( x − 5 ) 7 integral hisoblansin.
Yechish:
 13 dx
(x−5)7= 13  d(x−5)
(x−5)7=13  (x−5)−7∙d(x− 5)=¿
¿ 13 ∙ ( x − 5 ) − 6
− 6 + C = − 13
6 ( x − 5 ) 6 + C .
3.  5 x − 7
x 2
+ 3 x + 8 dx
integral hisoblansin.
Yechish:
 5 x − 7
x 2
+ 3 x + 8 dx =
 5
2
( 2 x + 3 ) + ( − 7 − 15
2 )
x 2
+ 3 x + 8 dx = 5
2  2 x + 3
x 2
+ 3 x + 8 dx − ¿ ¿
− 29
2  dx
x 2
+ 3 x + 8 = 5
2 ln
| x 2
+ 3 x + 8 | − 29
2  d ( x + 1,5 )
(
x + 1,5 ) 2
+ 5,75 = ¿ ¿
¿ 5
2 ln
| x 2
+ 3 x + 8 |
− 29
2 ∙ 1 √
5,75 arctg x + 1,5 √
5,75 + C = 5
2 ln
| x 2
+ 3 x + 8 | − ¿
− 29
√
23 arctg 2 x + 3 √
23 + C
.
4.
 (x−1)dx
(x¿¿2+2x+3)2¿ integral hisoblansin.
Yechish:
 (x−1)dx
(x¿¿2+2x+3)2=
1
2(2x+2)−2
(x¿¿2+2x+3)2dx
= 1
2 2x+2
(x¿¿2+2x+3)2dx −¿¿
¿¿¿
− 2
 dx
[ ( x + 1 ) ¿ ¿ 2 + 2 ] 2 = 1
2  ( x ¿ ¿ 2 + 2 x + 3 ) − 2
d ( x ¿ ¿ 2 + 2 x + 3 ) − ¿ ¿ ¿ ¿
−2 dx
[(x+1)¿¿2+(√2)2]2=(
x+1=t
dx = dt )= −1
2(x¿¿2+2x+3)−¿¿¿

−2 dt
(t2+2)2= − 1
2(x¿¿2+2x+3)− t2+2−t2
(t2+2)2dt =¿¿¿ − 1
2 ( x ¿ ¿ 2 + 2 x + 3 ) −
 t 2
+ 2
( t 2
+ 2 ) 2 dt +
 t 2
( t 2
+ 2 ) 2 dt = ¿ ¿
¿ − 1
2 ( x
¿ ¿ 2 + 2 x + 3 ) −
 dt
t 2
+ 2 +
 tdt
( t 2
+ 2 ) 2 ∙ t = − 1
2 ( x ¿ ¿ 2 + 2 x + 3 ) ¿ ¿
−1
√2arctg t
√2− t
2(t2+2)
+1
2 dt
t2+2
= −1
2(x¿¿2+2x+3)− ¿¿¿
−1
√2arctg t
√2− t
2(t2+2)
+ 1
2√2arctg t
√2=¿− 1
2(x¿¿2+2x+3)−¿¿
−1
√2arctg x+1
√2 − x+1
2(x¿¿2+2x+3)+ 1
2√2arctg x+1
√2 +C=¿¿
¿− x+2
2(x¿¿2+2x+3)− 1
2√2arctg x+1
√2 +C .¿
Mustaqil yechish uchun topshiriqlar:
1. Quyidagi integrallar hisoblansin.
1)  3 dx
x − 19 ; 2)  7 dx
x + 5 ; 3)  3,5 dx
x − 6,5 ;
4)  dx
x 2
+ 2 x + 5 ; 5)  dx
3 x 2
− 2 x + 4 ; 6)  dx
x 2
+ 3 x + 1 ;
7)  dx
x 2
− 6 x + 5 ; 8)  dx
2 x 2
− 2 x + 1 ; 9)  dx
3 x 2
− 2 x + 2 ;

10)  3 x − 1
5 x 2
− 3 x + 2 dx
; 11)  xdx
x 2
+ x + 1 ; 12)  2 x + 7
x 2
+ x − 2 dx
;
13)  x − 1
x 2
− 5 x + 6 dx
; 14)  4 x − 2,4
x 2
− 0,2 x + 0,17 dx
;
15)  ( 2 x + 1 ) dx
( x 2
+ 2 x + 5 ) 2 ; 16)  ( 3 x + 5 ) dx
( x 2
+ 2 x + 5 ) 2 .
Javob: 1) 3 ln| x − 19 | + C
; 2) 7 ln | x + 5 | + C
;
3) 3,5 ln
| x − 6,5 | + C
; 4) 1
2 arctg x + 1
2 + C
; 5) 1 √
11 arctg 3 x − 1 √
11 + C
;
6) 1
√
5 ln | 2 x + 3 −
√ 5
2 x + 3 +
√ 5 | + C
; 7) 1
4 ln | x − 5
x − 1 | + C
; 8) arctg ( 2 x − 1 ) + C
; 9)
1
√
5 arctg 3 x − 1 √
5 + C
; 10) 3
2 ln
| x 2
− x + 1 | + 1
√
3 arctg 2 x − 1 √
3 + C
;
11) 1
2 ln
( x 2
+ x + 1 ) − 1
√
3 arctg 2 x + 1 √
3 + C
;
12)
ln|
(x−1)3
x+2 |+C ; 13) ln|
c(x−2)2
x−3 |+C ;
14) 2 ln
( x 2
− 0,2 x + 0,17 ) − 5 arctg 10 x − 1
4 + C
;
15) – x − 9
8 ( x
¿ ¿ 2 + 2 x + 5 ) − 1
16 arctg x + 1
2 + C ¿ ;
16)
x−5
4(x¿¿2+2x+5)+1
8arctg x+1
2 +C ¿ .
Xulosa:

Yuqorida bayon etilganlarning hammasidan, har qanday ratsional
funksiyadan olingan integral ohirida elementar funksiyalar orqali ifoda
etilishi mumkin ekanligi kelib chiqadi, jumladan
1) I tipdagi sodda kasrlar bo’lgan holda – logarifmlar bilan;
2) II tipdagi sodda kasrlar bo’lgan holda – ratsional funksiyalar bilan;
3) III tipdagi sodda kasrlar bo’lgan holda – logarifmlar va arktangenslar bilan;
4) IV tipdagi sodda kaslar bo’lgan holda – ratsional funksiyalar va
arktangenslar bilan chekli shaklda ifoda etiladi.
Foydalanilgan adabiyotlar
2. Claudio Canuto, Mathematical Analysis I. 2008.

3. PETER V. O’NEIL, ADVANCED ENGINEERING MATHEMATICS.
2010.
4. Crowell and Slesnick, Calculus with Analytic Geometry. 2008
5. Algebra va matematik analiz asoslar. A.U.Abduhamidov va
boshqalar.Toshkent 2014-yil.
6. Aniqmas integrallarda integrallash usullari. A.Begmatov. Samarqand. 2011

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