Reja: Irratsional ifodalarni integrallash

2-misol. integralni toping.
Yechish:



> restart;
> with(student):
> IR11:=changevar(x+1=t^6,int(sqrt(x+1)/(1+(x+1)^(1/3))
,x),t);

> IR11:=changevar(t=(x+1)^(1/6), (IR11, t),x);

3-misol. integralni toping.
Bunda ,
=
1) o`zgaruvchini almashtirish yordamida integralni topish
> restart;
> with(student):
>IR13:=changevar(x=(1-t^2)/(1+t^2),int(sqrt((1-x)/(1+x))/
(1-x)^2, x),t);

> IR13:=changevar(t=sqrt((1-x)/(1+x)), (IR13, t),x);

2)Bevosita integrallash.
> restart;
> IR13:=Int(sqrt((1-x)/(1+x))/(1-x)^2,x)=int(sqrt((1-x)
/(1+x))/(1-x)^2,x);

2. integral binomial differensial integrali deb atalib, bu yerda m,n,p lar ratsional, a va b lar esa noldan farqli haqiqiy sonlardir. P.L.Chebishev tomonidan, bu integral :
1) p- butun son (bo`lganda yoyish yordami bilan);
2) -butun son (bo`lganda almashtrish bilan, s bunda p ni maxraji);
3) - butun son (bo`lganda almashtrish bilan, s bunda p ni maxraji) bo`lgan hollardan biri sodir bo`lgandagina elementar funksiyadan iborat bo`lishi, ya`ni olinishi isbotlangandir. Boshqa holda u olinmaydigan integraldir.
4-misol. integralni toping.
Yechish. Bunda 2) hol bajarilishidan
, ,
almashtrish bilan

=
1) o`zgaruvchini almashtirish yordamida integralni topish
> restart;
> with(student):
> IR14:=changevar(x=(t^3-1)^4,int((1+x^(1/4))^(1/3)/sqrt(x),x),t);

> IR14:=changevar(t=(1+x^(1/4))^(1/3), (IR14, t),x);