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Index stranica MATEMATIČKA ANALIZA INTEGRALI

Integrali racionalnih funkcija

[inlmath]\int xe^x\mathrm dx[/inlmath]

Integrali racionalnih funkcija

Postod slavonija035 » Četvrtak, 25. April 2013, 10:01

[dispmath]602.\quad\int\frac{x^4+1}{x^3-x}\mathrm dx[/dispmath][dispmath]604.\quad\int\frac{x^2+2x}{\left(x+1\right)^4}\mathrm dx[/dispmath][dispmath]608.\quad\int\frac{x\mathrm dx}{x^3-1}[/dispmath][dispmath]609.\quad\int\frac{\mathrm dx}{x^3+x^2+x+1}[/dispmath]
ja sam ovo sve pokušao riješiti rastavljanjem funkcije na parcijalne razlomke, ali nisam uspio dobiti točna rješenja.
postoji li neka druga metoda koja može biti jednostavnija kod rješavanja ovakvih i sličnih integrala?

P.S. nisam siguran jel ovo trebalo bit u novoj temi :)
Poslednji put menjao Daniel dana Četvrtak, 10. April 2014, 12:45, izmenjena 2 puta
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Re: Integrali racionalnih funkcija

Postod Daniel » Četvrtak, 25. April 2013, 13:28

Da, sasvim je ok nova tema, bez brige. :)

Ja zaista ne vidim drugog načina, bar ne tako jednostavnog, kao preko parcijalnih razlomaka. Evo ti zasad postupci za prva dva, da možeš da uporediš sa svojim postupcima i pronađeš eventualnu grešku.

slavonija035 je napisao:[dispmath]602.\quad\int\frac{x^4+1}{x^3-x}\mathrm dx[/dispmath]

Pošto je stepen polinoma u brojiocu veći od stepena polinoma u imeniocu, treba prvo da ih podelimo:

[inlmath]\left(x^4+1\right):\left(x^3-x\right)=x\\
x^4-x^2\\
\phantom{x^4-}x^2+1[/inlmath]

pa integral možemo pisati kao:
[dispmath]\int\left(x+\frac{x^2+1}{x^3-x}\right)\mathrm dx=\int x\mathrm dx+\int\frac{x^2+1}{x^3-x}\mathrm dx=\frac{x^2}{2}+\int\frac{x^2+1}{x^3-x}\mathrm dx[/dispmath]
E sad, ovaj razlomak u drugom integralu, [inlmath]\frac{x^2+1}{x^3-x}[/inlmath], rastavimo na parcijalne razlomke:
[dispmath]\frac{x^2+1}{x^3-x}=\frac{x^2+1}{x\left(x^2-1\right)}=\frac{x^2+1}{x\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}=\frac{A\left(x^2-1\right)+Bx\left(x+1\right)+Cx\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=[/dispmath][dispmath]=\frac{Ax^2-A+Bx^2+Bx+Cx^2-Cx}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(A+B+C\right)x^2+\left(B-C\right)x-A}{x\left(x-1\right)\left(x+1\right)}[/dispmath]
Odatle imamo sistem:
[inlmath]\begin{array}{ll}
A+B+C=1\\
B-C=0\\
-A=1
\end{array}[/inlmath]

a rešenja su: [inlmath]A=-1,\;B=1,\;C=1[/inlmath]

pa je
[dispmath]\frac{x^2+1}{x^3-x}=-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x+1}[/dispmath]
a integral je
[dispmath]\frac{x^2}{2}+\int\frac{x^2+1}{x^3-x}\mathrm dx=\frac{x^2}{2}+\int\left(-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right)\mathrm dx=\frac{x^2}{2}-\int\frac{\mathrm dx}{x}+\int\frac{\mathrm dx}{x-1}+\int\frac{\mathrm dx}{x+1}=[/dispmath][dispmath]=\frac{x^2}{2}-\ln\left|x\right|+\ln\left|x-1\right|+\ln\left|x+1\right|+c=\frac{x^2}{2}+\ln\left|c\frac{\left(x-1\right)\left(x+1\right)}{x}\right|=\frac{x^2}{2}+\ln\left|c\frac{x^2-1}{x}\right|[/dispmath]
slavonija035 je napisao:[dispmath]604.\quad\int\frac{x^2+2x}{\left(x+1\right)^4}\mathrm dx[/dispmath]

Ovde je stepen polinoma u brojiocu manji od stepena polinoma u imeniocu, pa odmah možemo krenuti s rastavljanjem na parcijalne razlomke:
[dispmath]\frac{x^2+2x}{\left(x+1\right)^4}=\frac{A}{x+1}+\frac{B}{\left(x+1\right)^2}+\frac{C}{\left(x+1\right)^3}+\frac{D}{\left(x+1\right)^4}=\frac{A\left(x+1\right)^3+B\left(x+1\right)^2+C\left(x+1\right)+D}{\left(x+1\right)^4}=[/dispmath][dispmath]=\frac{A\left(x^3+3x^2+3x+1\right)+B\left(x^2+2x+1\right)+Cx+C+D}{\left(x+1\right)^4}[/dispmath]
Da ne bismo sve računali, odmah možemo da primetimo da je [inlmath]A=0[/inlmath], jer se jedino uz [inlmath]A[/inlmath] nalazi kubni član, kojeg nema u polinomu iz brojioca. Izraz postaje:
[dispmath]\frac{B\left(x^2+2x+1\right)+Cx+C+D}{\left(x+1\right)^4}[/dispmath]
Pošto se jedino uz [inlmath]B[/inlmath] nalazi kvadratni član, a znamo da je u polinomu iz brojioca koeficijent uz kvadratni član jednak [inlmath]1[/inlmath], biće [inlmath]B=1[/inlmath]:
[dispmath]\frac{\left(x^2+2x+1\right)+Cx+C+D}{\left(x+1\right)^4}=\frac{x^2+\left(C+2\right)x+\left(C+D+1\right)}{\left(x+1\right)^4}[/dispmath]
i odatle imamo
[inlmath]\begin{array}{ll}
C+2=2\\
C+D+1=0
\end{array}[/inlmath]

[inlmath]C=0,\;D=-1[/inlmath]

pa je razlomak jednak
[dispmath]\frac{x^2+2x}{\left(x+1\right)^4}=\frac{1}{\left(x+1\right)^2}-\frac{1}{\left(x+1\right)^4}[/dispmath]
a integral je
[dispmath]\int\frac{x^2+2x}{\left(x+1\right)^4}\mathrm dx=\int\left[\frac{1}{\left(x+1\right)^2}-\frac{1}{\left(x+1\right)^4}\right]\mathrm dx=\int\frac{\mathrm dx}{\left(x+1\right)^2}-\int\frac{\mathrm dx}{\left(x+1\right)^4}=[/dispmath][dispmath]=-\frac{1}{x+1}+\frac{1}{3\left(x+1\right)^3}+c=\frac{1-3\left(x+1\right)^2}{3\left(x+1\right)^3}+c=-\frac{3x^2+6x+2}{3\left(x+1\right)^3}+c[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
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Re: Integrali racionalnih funkcija

Postod Daniel » Četvrtak, 25. April 2013, 15:12

slavonija035 je napisao:[dispmath]608.\quad\int\frac{x\mathrm dx}{x^3-1}[/dispmath]

[dispmath]\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}=\frac{A\left(x^2+x+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=[/dispmath][dispmath]=\frac{Ax^2+Ax+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(A+B\right)x^2+\left(A-B+C\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+x+1\right)}[/dispmath]
[inlmath]\begin{array}{ll}
A+B=0\\
A-B+C=1\\
A-C=0
\end{array}[/inlmath]

[inlmath]\Rightarrow\quad A=\frac{1}{3},\;B=-\frac{1}{3},\;C=\frac{1}{3}[/inlmath]
[dispmath]\frac{x}{x^3-1}=\frac{\frac{1}{3}}{x-1}+\frac{-\frac{1}{3}x+\frac{1}{3}}{x^2+x+1}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{x-1}{x^2+x+1}\right)[/dispmath]
[dispmath]\int\frac{x\mathrm dx}{x^3-1}=\frac{1}{3}\int\left(\frac{1}{x-1}-\frac{x-1}{x^2+x+1}\right)\mathrm dx=\frac{1}{3}\int\frac{\mathrm dx}{x-1}-\frac{1}{3}\int\frac{x-1}{x^2+x+1}\mathrm dx=[/dispmath][dispmath]=\frac{1}{3}\ln\left|x-1\right|-\frac{1}{6}\int\frac{2x-2}{x^2+x+1}\mathrm dx=\frac{1}{3}\ln\left|x-1\right|-\frac{1}{6}\int\frac{2x+1-3}{x^2+x+1}\mathrm dx=[/dispmath][dispmath]=\frac{1}{3}\ln\left|x-1\right|-\frac{1}{6}\int\frac{2x+1}{x^2+x+1}\mathrm dx+\frac{1}{6}\int\frac{3}{x^2+x+1}\mathrm dx=[/dispmath][dispmath]=\frac{1}{3}\ln\left|x-1\right|-\frac{1}{6}\int\frac{\mathrm d\left(x^2+x+1\right)}{x^2+x+1}+\frac{1}{2}\int\frac{\mathrm dx}{x^2+x+1}=[/dispmath][dispmath]=\frac{1}{3}\ln\left|x-1\right|-\frac{1}{6}\ln\left|x^2+x+1\right|+\frac{1}{2}\int\frac{\mathrm dx}{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}+1}=[/dispmath][dispmath]=\frac{1}{6}\ln\left(x-1\right)^2-\frac{1}{6}\ln\left|x^2+x+1\right|+\frac{1}{2}\int\frac{\mathrm dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}=[/dispmath][dispmath]=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{4}{2}\int\frac{\mathrm dx}{4\left(x+\frac{1}{2}\right)^2+3}=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{4}{2}\int\frac{\mathrm dx}{\left(2x+1\right)^2+3}=[/dispmath][dispmath]=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+2\int\frac{\mathrm dx}{3\left[\left(\frac{2}{\sqrt 3}x+\frac{1}{\sqrt 3}\right)^2+1\right]}=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{2}{3}\int\frac{\frac{\sqrt 3}{2}\mathrm d\left(\frac{2}{\sqrt 3}x\right)}{\left(\frac{2}{\sqrt 3}x+\frac{1}{\sqrt 3}\right)^2+1}=[/dispmath][dispmath]=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{\sqrt 3}{3}\int\frac{\mathrm d\left(\frac{2}{\sqrt 3}x+\frac{1}{\sqrt 3}\right)}{\left(\frac{2}{\sqrt 3}x+\frac{1}{\sqrt 3}\right)^2+1}=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{\sqrt 3}{3}\mathrm{arctg}\left(\frac{2}{\sqrt 3}x+\frac{1}{\sqrt 3}\right)+c=[/dispmath][dispmath]=\frac{1}{6}\ln\frac{\left(x-1\right)^2}{\left|x^2+x+1\right|}+\frac{\sqrt 3}{3}\mathrm{arctg}\left[\frac{\sqrt 3\left(2x+1\right)}{3}\right]+c[/dispmath]
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Re: Integrali racionalnih funkcija

Postod Daniel » Petak, 26. April 2013, 01:25

slavonija035 je napisao:[dispmath]609.\quad\int\frac{\mathrm dx}{x^3+x^2+x+1}[/dispmath]

[dispmath]\frac{1}{x^3+x^2+x+1}=\frac{1}{x^2\left(x+1\right)+\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}=[/dispmath][dispmath]=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=[/dispmath][dispmath]=\frac{Ax^2+A+Bx^2+Bx+Cx+C}{\left(x+1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(B+C\right)x+\left(A+C\right)}{\left(x+1\right)\left(x^2+1\right)}[/dispmath]
[inlmath]\begin{array}{ll}
A+B=0\\
B+C=0\\
A+C=1
\end{array}[/inlmath]

[inlmath]\Rightarrow\quad A=\frac{1}{2},\;B=-\frac{1}{2},\;C=\frac{1}{2}[/inlmath]
[dispmath]\frac{1}{x^3+x^2+x+1}=\frac{\frac{1}{2}}{x+1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{x-1}{x^2+1}\right)[/dispmath]
[dispmath]\int\frac{\mathrm dx}{x^3+x^2+x+1}=\frac{1}{2}\int\left(\frac{1}{x+1}-\frac{x-1}{x^2+1}\right)\mathrm dx=\frac{1}{2}\int\frac{\mathrm dx}{x+1}-\frac{1}{2}\int\frac{x-1}{x^2+1}\mathrm dx=[/dispmath][dispmath]=\frac{1}{2}\ln\left|x+1\right|-\frac{1}{2}\left(\int\frac{x}{x^2+1}\mathrm dx-\int\frac{\mathrm dx}{x^2+1}\right)=\frac{1}{2}\ln\left|x+1\right|-\frac{1}{2}\cdot\frac{1}{2}\int\frac{\mathrm d\left(x^2\right)}{x^2+1}+\frac{1}{2}\mathrm{arctg}\:x=[/dispmath][dispmath]=\frac{1}{2}\ln\left|x+1\right|-\frac{1}{4}\ln\left|x^2+1\right|+\frac{1}{2}\mathrm{arctg}\:x+c=\frac{1}{4}\ln\left(x+1\right)^2-\frac{1}{4}\ln\left|x^2+1\right|+\frac{1}{2}\mathrm{arctg}\:x+c=[/dispmath][dispmath]=\frac{1}{4}\ln\frac{\left(x+1\right)^2}{\left|x^2+1\right|}+\frac{1}{2}\mathrm{arctg}\:x+c[/dispmath]
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