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Bo`laklab integrallash usuli. Faraz qilaylik,
u(x)
va v(x)
funksiyalar uzluksiz

u^(x) ,
v^(x)
hosilalarga ega bo`lsin.

Ravshanki,
bo`ladi. Demak,
funksiya

(u(x)  v(x))^  u^(x)  v(x)  u(x)  v^(x)

F (x)  u(x)  v(x)


f (x)  u^(x)  v(x)  u(x)  v^(x)

funksiyaning boshlang`ich funksiyasi bo`ladi. Bundan

_u^(x)  v(x)  u(x)  v^(x)dx  u(x)  v(x)  C

bo`lishi kelib chiqadi.
Aniqmas integralning 3)- va 4)- xossalardan foyda-lanib
<sub></sub> u(x)  v<sup></sup>(x)dx u(x)  v(x)  <sub></sub> u<sup></sup>(x)  v(x)dx
bo`lishini topamiz.
(5) formulani quyidagicha

(5)

ham yozish mumkin.
_u(x)  dv(x) u(x)  v(x)  _ v(x)du(x)
(5‰)

Bu (5‰) formula bo`laklab integrallash formulasi deyiladi. Uning yordamida

integralni hisoblash
integralni hisoblashga keltiriladi.

5. misol.

integral hisoblansin.
_ u(x)  v^(x)dx


_ u^(x)  v(x)dx
_ x cos xdx

◄Bo`laklab integrallash formulasidan foydalanib topamiz:

u  x ,
^{ x cos xdx } cos xdx  dv
^{du  dx}  x sin x  sin xdx 


v  sin x

6. 
   misol. Ushbu
   

 x sin x  cos x  C. ►

integral hisoblansin.
◄ Qaralayotgan integralda
J  _
x²  adx

deyilsa, unda
u  ,
dv  dx

du 
^x dx ,


v  x

bo`ladi. Bo`laklab integrallash formulasidan foydalanib topamiz:
_x2

J  x

 x

• 
  _ dx  x x²  a
  

 _
 _
dx  a_ ^dx 
dx 

 x  J  a_ ^dx .

Demak,

J  x

• 
  J  a_ ^dx ,
  

₂ 
J  ^{1 }x


• 
  a ^{dx } .
  

Ma`lumki, (1⁰ dagi 4-misol)
_ ^dx  ln x 

• 
  C .
  

Natijada


J  _


dx  ^x
2

• 
  ^a ln x   C
  

2

bo`lishi kelib chiqadi. ►

7. 
   misol. Ushbu
   

J_n 

integral topilsin.

◄ Bu integralda
dx




(x ²  a ² )ⁿ


u  ¹
(n  N , a  R, a  0)
, dv  dx

deb olsak, unda

(x ²  a ² )ⁿ

du  
2nxdx
_(x2 _{ a}2 ₎n1 ^,


v  x

bo`ladi. (5) formuladan foydalanib topamiz:
x x ²

J 
ⁿ (x ²  a ² )ⁿ
 2n_{ (x 2  a 2 )n1} dx 


 ^x  2n^

dx _{ a 2}

dx ^


.

Natijada
(x²  a ² )ⁿ

^ (x²  a ² )^n
 _(x2 _{ a} 2 ₎n1 ^

J_n 
x
(x ²  a ² )ⁿ

• 
  2n  J_n
  

• 
  2na²  J
  

n1

bo`ladi. Bu tenglikdan

^Jn1
_ 1
2na ²
x
(x ²  a ² )ⁿ
_ 2n 1 1
²ⁿ a ²
 J_n

(6)

bo`lishi kelib chiqadi. ►

Odatda, (6) munosabat rekkurent formula deyiladi.

Ravshanki,
n  1 bo`lganda


dx
x
d ( )
1 _a 1 x

bo`ladi.
^{J1  } x ²  a ²
 _{a }
1  ( ^x )²
a
 arctg  C
a a

n  2 bo`lganda mos topiladi.
Masalan,
J _n integrallar (6) rekkurent formula yordamida

_{J } dx _ 1

x _ 1
 J  ^{1 x}  ¹

arctg ^x  C

bo`ladi.
^{2 } (x²  a² ) ² 2a²

x²  a² 2a^2
1 2a² (x²  a² )
2a^{3 a}

4-§. Sodda kasrlarni integrallash

Ushbu


A
(x  a)^m
(x  a) ,


Bx  C


(x ²  px  q)^m

ko`rinishdagi funksiyalar sodda kasrlar deyiladi, bunda
m  N;
A, B,C , a , p , q –

haqiqiy sonlar bo`lib, x<sup>2</sup>  px  q kvadrat uchhad haqiqiy ildizga ega emas, ya`ni

_p 2
q   0 .
4


m  1 bo`lganda sodda kasrlarning integrallari

lar quyidagicha hisoblanadi:

^A dx ,


x  a
Bx  C _dx x ²  px  q

_ A _{dx  A} d (x  a) _{ A ln}
x  a  C ;

x  a x  a
_ Bx  C _{dx  } Bx  C _{dx }

x²  px  q
(x 
p ₎₂
2
 (q  ^p )
4

x  ^p  t,
_ 2
x  t  ^p
2 _
p 2 ₂

dx  dt,
q   a
4

_{ B} tdt

• 
  (C  ^Bp ) ^dt  ^B ln(t ²  a² )  (C  ^{Bp 1}
  

^t  C* 

^ t ²  a²
2 ^ t ²  a² 2
) arctg
2 a a

 ^B ln(x²  px  q) 
2
2C  Bp
arctg
x  ^p
2
 C *.

Aytaylik,
m  N , m  1
bo`lsin. Bu holda sodda asrlarning integrallari


^A dx , (x  a)^m
lar quyidagicha hisoblanadi:
Bx  C _dx


(x ²  px  q)^m

_ ^A dx  A_ (x  a)^m d (x  a)   ^A  C,

(x  a)^m
(m  1)( x  a)^m1
x  ^p  t, x  t  ^p

 ^{2 m}
Bx  C _{dx }
(x  px  q)
2 2 _
p2

_ В _ 2tdt _{ (C }
dx  dt,


^p B)_
q   a²
4
dt _

² (t ²  a ² )^{m 2} (t ²  a ² )^m

  ^{B 1}  (C  ^p B)_ ^dt .

Keyingi munosabatdagi

² (m  1)(t ²  a ² )^{m1 2}


dt _. (t ²  a ² )^m
(t ²  a ² )^m

integral (6 ) rekurrent formula yordamida topiladi.

1. 
   Ushbu
   

_ dx
Mashqlar

integral hisoblansin.

2. 
   Ushbu