Bizga ikkita diferensiallanuvchi u(X) va v(X) funksiyalar berilgan

 ... 

(x  b)^k
x²  px  q
(x²  px  q)²

 ... 

M _n x  N_n

 ... 

C₁ x  D₁

 ... 

C_r x  D_r


(*)

(x ²  px  q) ⁿ
x ²  cx  d
(x ²  cx  d ) ^r

bu yerda A_m,...,B_k,...,M_n,...N_n,...,C_r,...,D_r lar o’zgarmas sonlar. Q(x) necha karrali ildizga ega bo’lsa, sxemadagi elementar kasrlar soni shuncha bo’ladi.

v) Hosil bo’lgan (*) tenglikni har ikki tomonini Q(x) ga ko’paytirish bilan kasrni maxrajdan qutqaramiz.
g) Кeyin hosil bo’lgan tenglikni har ikki tomonidagi x ning bir xil darajalari oldidagi koeffitsientlarni tenglashtirib, tenglamalar sistemasini hosil qilamiz.
Bu sistemadagi tenglamalar soni A₁, ..., B₁, ..., M₁, ..., N₁, ..., C₁,..., D₁, ..., noma’lumlar soniga teng bo’lishi kerak.
d) Hosil bo’lgan tenglamalar sistemasi yechilib, noma’lum koeffitsientlar topiladi va ular (*) ayniyatga qo’yiladi va ikki tomoni dx ga ko’paytirilib, integrallanadi. Hosil bo’lgan elementar kasrlar (1)-(4) ko’rinishdagi kasrlardan iborat bo’ladi.
3x ²  8

Miso l.1.
^ x ³  4x ²  4x^dx
hisoblansin.

Yechish. Integral ostidagi kasrning maxrajini ko’paytuvchilarga ajratamiz.
x³+4x²+4x=x(x²+4x+4)=x(x+2)²
Endi berilgan kasrni (*) dan foydalanib, elementar kasrlarga yoyamiz:

3x²  8

x(x  2)²
 ^A 

x

B _
x  2
C
(x  2)²

(**)

buni har ikki tomonini x(x+2)² ga ko’paytiramiz.
3x²+8=A(x+2)²+Bx(x+2)+Cx=(A+B)x²+(4A+2B+C)x+4A
Endi x ni bir xil darajalari oldidagi koeffitsientlarini tenglashtirib, tenglamalar sistemasini tuzamiz:
^ A  B  3



_4 A  2B  C  0


^ 4 A  8
Bu sistemani yechib, A=2, B=1, C=-10 larni topamiz. So’ngra bularni (**) ga qo’yamiz.

3x ²  8 2 1 10

x(x  2)^{2 } x ^ x  2 ^ (x  2)²
Buni ikki tomonini dx ga ko’paytirib, keyin integrallaymiz:

3x ²  8
2 1 10

^ x³  4x ²  4x^{dx  [ x  x  2 } (x  2)^{2 ]dx }
 2_ ^dx  _ ^dx 10_ (x  2)^2 d (x  2) 

x
x ³  4x ²
x  2
=2ln|x|+ln|x+2|+10/(x+2)+C
 2x  1

Misol 2. 
x ⁴  x ^dx
hisoblansin.

Yechish. Berilgan kasrni elementar kasrlarga ajratamiz. Buning uchun maxrajdagi ko’phadni ko’paytuvchilarga ajratamiz
x⁴+x=x(x³+1)=x(x+1)(x²-x+1)

x ³  4x ²  2x  1 A B
Cx  D

x ⁴  x ^ x ^ x  1 ^ x ²  x  1
x³+4x²-2x+1=A(x³+1)+Bx(x²-x+1)+(Cx+D)(x²+x)=
=(A+B+C)x³+(C+D-B)x²+(B+D)x+A
Endi x larning bir xil darajalari oldidagi koeffitsientlarni tenglashtirish bilan noma’lum A, B, C, D larni aniqlash uchun qo’yidagi 4 ta tenglamani hosil qilamiz, hamda bu tenglamalar sistemasini yechib, A, B, C, D noma’lumlarni aniqlaymiz:

_ A  B  C  1
_C  D  B  4
A  1
B  2


^ B  D  2
_^ A  1
C  2
D  0

Topilganlarni noma’lumlar o’rniga qo’yib kasrni elementar kasrlar orqali ifodasini yozamiz:
x ³  4x ²  2x  1 1 2 2x
x ⁴  x ^ x ^ x  1 ^ x ²  x  1


endi buni integrallaymiz.

J  _
x ³  4x ²
_x 4
 2x  1

• _x dx 

dx _{ 2}


x

dx _{ 2}


x  1
xdx _
x ²  x  1

=ln|x|-2ln|x+1|+2J₁


J₁ integralda x²-x+1 dan to’la kvadrat ajratamiz:
x²-x+1=(x-1/2)²+3/4 Bunda x-1/2=t dx=dt deb olamiz. U holda J₁ qo’yidagicha hisoblanadi.

(t  1 / 2)dt 1
J  
2tdt _ 1


dt _
¹ ln(t ²

 3 / 4) 

1  _t 2

 3 / 4

₂  _t 2

 3 / 4

₂  _t 2

 3 / 4 2

• 
  ¹ arctg ^2t
  

 ¹ ln(x ²

2

 x  1) 
1 2x  1

arctg .

Natijada J integral qo’yidagicha aniqlanadi:
| x|(x ²  x  1) 2


2x  1

J  ln
(x  1)^{2 }
arctg  C

(x ³  3)dx

Miso l.3.
 _x 4
 10x ²
 25
hisoblansin.

Yechish. Maxrajni ko’paytuvchilarga ajratamiz:

x⁴ 10x²
 25  (x²
 5)²

x³  3 _
x⁴ 10x²  25
Ax  B _
x²  5
Cx  D


(x²  5)²

x³  3  (Ax  B)(x²  5)  Cx  D  Ax³  Bx²  (5A  C)x  (5B  D)

 ^{A  1}


^ B  0
A  1
B  0


^ 5A  C  0
C  5

^_5B  D  3 D  3

Buni integrallaymiz:

x³  3 _
x⁴ 10x²  25
x _
x²  5
5x  3


(x²  5)²

x³  3
xdx
xdx dx

J  _
x ⁴  10x ²  25 ^{dx  } x ²  5 ^{ 5} (x ²  5)^{2
 3} (x ²  5)^{2 .}

––
J₁
––
J ₂
––
J₃

xdx
1 2xdx
1 d (x ²  5) 1 ₂

J₁  _
x ²  5 ^{ 2 }
x ²  5 ^{ 2 }
x ²  5
 ln(x
2
 5)

xdx
¹ 2 2 2
1 (x ²  5)^1 1

J ₂  _
(x ²  5)²
 _{2 } (x
 5)
d (x
 5) 
2
 
^1 2(x ²  5)

J₃ integralda o’zgaruvchini almashtiramiz:

x  5tgzdx 
dx
5 sec² zdz
5 sec² zdz 1 ₂

J₃  _
(x²  5)^{2  }

25sec⁴ z
_ cos
zdz 

_ 1 _
(1 cos 2z)dz  ¹
(z  ¹ sin 2z) 
2
¹ (arctg
^x  ).
x²  5

Berilgan integral qo’yidagiga teng bo’ladi:

J  ¹ ln( x ²  5) 
5 _ 3
(arctg
x _ x
⁵ )  C 

2 2(x ²  5) x ²  5

 ¹ ln( x ²  5) 
2
25  3x _
10(x ²  5)
³ arctg
^x  C.