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səhifə | 4/10 | tarix | 10.11.2022 | ölçüsü | 0,96 Mb. | | #68465 |
| soddaBu səhifədəki naviqasiya:
- 2) > A:=4:B:=5:p:=6:q:=25:a:=sqrt(4*q-p^2)/2: > with(student): > changevar(t=x+p/2, (Ik3, t), x); 3) n=2,3 bo`lganda bevosita topish.
- > restart; > f:=x->(1-x^3)/(x^5+x^2); > y1:=x->A1/x+A2/x^2+A3/(x+1)+(A4*x+A5)/(x^2-x+1); > p:=simplify(y1(x));
- > kp2:=coeff(pol1,x,2)=coeff(pol2,x,2); > kp3:=coeff(pol1,x,3)=coeff(pol2,x,3); > kp4:=coeff(pol1,x,4)=coeff(pol2,x,4);
- > int(y1(x),x);
> m:=3:value(Ik2);
> m:=4:Ik2:=value(Ik2);
> m:=4:a:=1:Ik2:=value(Ik2);
n=2,3 bo`lganda integralini formula asosida topish.
1) > restart;
> Ik3:=int((A*(t-p/2)+B)/(t^2+a^2)^m,t);
> m:=2:Ik3:=value(Ik3);
> m:=3:Ik3:=value(Ik3);
2) > A:=4:B:=5:p:=6:q:=25:a:=sqrt(4*q-p^2)/2:
> with(student):
> changevar(t=x+p/2, (Ik3, t), x);
3) n=2,3 bo`lganda bevosita topish.
> restart;
> Ik4:=int((4*x+5)/(x^2+6*x+25)^2,x);
> Ik5:=int((4*x+5)/(x^2+6*x+25)^3,x);
3-misol. integralni toping.
Yechish:
hosil bolgan tenglikning chap va o`ng ko`phadlarining darajalari bo`icha mos koeffitsientlarni tenglashtirib quyidagi sistemani tuzamiz:
Sodda kasirlarning koeffitsientlarni topish va integrallash:
> restart;
> f:=x->(1-x^3)/(x^5+x^2);
> y1:=x->A1/x+A2/x^2+A3/(x+1)+(A4*x+A5)/(x^2-x+1);
> p:=simplify(y1(x));
> pol1:=f(x)*(x^5+x^2); pol2:=p*x^2*(x+1)*(x^2-x+1);
> kp0:=coeff(pol1,x,0)=coeff(pol2,x,0);
> kp1:=coeff(pol1,x,1)=coeff(pol2,x,1);
> kp2:=coeff(pol1,x,2)=coeff(pol2,x,2);
> kp3:=coeff(pol1,x,3)=coeff(pol2,x,3);
> kp4:=coeff(pol1,x,4)=coeff(pol2,x,4);
> k:=solve({kp0,kp1,kp2,kp3,kp4},{A1,A2,A3,A4,A5});
> A1:=0; A4:=-2/3; A3:=2/3; A5:=-2/3; A2:=1;y1(x);
> int(y1(x),x);
Topilgan koeffitsientlar asosida berilgan kasirni sodda kasirlarga ajratilgan ko`rinishin yozamiz:
Bebosita sodda kasirlarga ajratish:
Dostları ilə paylaş: |
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