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Całkowanie metodą Ostrogradskiego

Całka wymierna Całkowanie metodą Ostrogradskiego Całka Całka nieoznaczona Rachunek całkowy

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#21 Jarekzulus

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Napisano 13.08.2015 - 00:22

c)\, \int \frac{x^2+x+1}{x^5-2x^4+x^3}dx  Metodą Ostrogradskiego, rozkład w poście 5

 

\int \frac{x^2+x+1}{x^5-2x^4+x^3}dx=\frac{Ax^2+Bx+C}{x^2(x-1)}+\int \frac{Dx+E}{x(x-1)}dx

 

\frac{x^2+x+1}{x^5-2x^4+x^3}=\frac{(2Ax+B)(x^3-x^2)-(Ax^2+Bx+C)(3x^2-2x)}{x^4(x-1)^2}+\frac{Dx+E}{x(x-1)}

 

\frac{x\cdot (x^2+x+1)}{x\cdot (x^3\left(x-1\right)^2)}=\frac{(2Ax+B)(x^3-x^2)-(Ax^2+Bx+C)(3x^2-2x)}{x^4(x-1)^2}+\frac{x^3(x-1)(Dx+E)}{x^4(x-1)^2}

 

x^2+x+1=\left(2Ax+B\right)\left(x^2-x\right)-\left(Ax^2+Bx+C\right)\left(3x-2\right)+\left(Dx+E\right)\left(x^3-x^2\right)

 

x^2+x+1=\left(2Ax^3+Bx^2-2Ax^2-Bx\right)-\left(3Ax^3+3Bx^2+3Cx-2Ax^2-2Bx-2C\right)+\left(Dx^4-Dx^3+Ex^3-Ex^2\right)

 

x^2+x+1=Dx^4+\left(-D+E-A\right)x^3+\left(-E-2B\right)x^2+\left(B-3C\right)x+2C

 

\{D=0\\-A-D+E=0\\-2B-E=1\\B-3C=1\\2C=1\Rightarrow \{D=0\\E=A\\-2B-E=1\\B-3C=1\\2C=1 \Rightarrow \{A=-6\\B=\frac{5}{2}\\C=\frac{1}{2}\\D=0\\E=-6

 

\int \frac{x^2+x+1}{x^5-2x^4+x^3}dx=\frac{-6x^2+\frac{5}{2}x+\frac{1}{2}}{x^2(x-1)}+\int \frac{-6}{x(x-1)}dx

 

\int \frac{x^2+x+1}{x^5-2x^4+x^3}dx=\frac{-6x^2+\frac{5}{2}x+\frac{1}{2}}{x^2(x-1)}-6\left(\ln \left(x-1\right)-\ln \left(x\right)\right)+C


Użytkownik Jarekzulus edytował ten post 13.08.2015 - 00:29

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:wave: :wave: :wave: Jeśli rzuciłem choć promyczek światła na problem który postawiłeś - podziękuj. pre_1433974176__syg.jpgNad kreską


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Napisano 25.09.2011 - 17:55

#22 Jarekzulus

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Napisano 13.08.2015 - 00:27

b)\int{\frac{4x^2-8x}{\left(x-1\right)^2\left(x^2+1\right)^2}dx  Przez części obliczono w poście 3

 

R=\left(x-1\right)^2\left(x^2+1\right)^2

 

R'=2\left(x-1\right)\left(x^2+1\right)^2+4\left(x^3+x\right)\left(x-1\right)^2=2\left(x-1\right)\left(x^2+1\right)\left(x\left(3x-2\right)+1\right)=(x-1)(x^2+1)(6x^2-4x+2)

 

\int{\frac{4x^2-8x}{\left(x-1\right)^2\left(x^2+1\right)^2}dx=\frac{Ax^2+Bx+C}{(x-1)(x^2+1)}+\int\frac{Dx^2+Ex+F}{(x-1)(x^2+1)}dx

 

\frac{4x^2-8x}{\left(x-1\right)^2\left(x^2+1\right)^2}=\frac{\left(2Ax+B\right)\left(x-1\right)\left(x^2+1\right)-\(x^2+1+2x(x-1)\) \left(Ax^2+Bx+C\right)}{\left(\left(x-1\right)\left(x^2+1\right)\right)^2}+\frac{Dx^2+Ex+F}{(x-1)(x^2+1)}

 

\frac{4x^2-8x}{\left(x-1\right)^2\left(x^2+1\right)^2}=\frac{\left(2Ax+B\right)\left(x-1\right)\left(x^2+1\right)-\(x^2+1+2x(x-1)\) \left(Ax^2+Bx+C\right)}{\left(x-1\right)^2\left(x^2+1\right)^2}+\frac{(Dx^2+Ex+F)(x-1)(x^2+1)}{(x-1)^2(x^2+1)^2}

 

\left(2Ax+B\right)\left(x-1\right)\left(x^2+1\right)-\(x^2+1+2x(x-1)\) \left(Ax^2+Bx+C\right)+(Dx^2+Ex+F)(x-1)(x^2+1)=4x^2-8x

 

Dx^5-Dx^4-Ax^4+Ex^4-Ex^3+Fx^3-2Bx^3+Dx^3-Dx^2-Fx^2-3Cx^2+Ex^2+Bx^2+Ax^2+Fx-Ex-2Ax+2Cx-B-C-F=4x^2-8x

 

\{D=0\\-D-A+E=0\\-E+F-2B+D=0\\-D-F-3C+E+B+A=4\\F-E-2A+2C=-8\\-B-C-F=0        \Rightarrow       \{A=3\\B=-1\\C=0\\D=0\\E=3\\F=1

 

więc

 

\int{\frac{4x^2-8x}{\left(x-1\right)^2\left(x^2+1\right)^2}dx=\frac{3x^2-1x}{(x-1)(x^2+1)}+\int\frac{3x+1}{(x-1)(x^2+1)}dx

 

\bl\int\frac{3x+1}{(x-1)(x^2+1)}dx=\int\(\frac{-2x+1}{x^2+1}+\frac{2}{x-1}\)dx=-\int\frac{2x}{x^2+1}dx+\int\frac{dx}{x^2+1}-\int\frac{2}{x-1}dx=-ln|x^2+1|+ arctg(x)+2ln|x-1|+C

 

Czyli

 

\int{\frac{4x^2+8x}{\left(x-1\right)^2\left(x^2+1\right)^2}dx=\frac{3x^2-x}{(x-1)(x^2+1)}-ln|x^2+1|+arctg(x)+2ln|x-1|+C


Użytkownik Jarekzulus edytował ten post 16.08.2015 - 16:36

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:wave: :wave: :wave: Jeśli rzuciłem choć promyczek światła na problem który postawiłeś - podziękuj. pre_1433974176__syg.jpgNad kreską


#23 Mariusz M

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Napisano 13.08.2015 - 13:36

g)        Przez części w poście 8

 

\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}

 

R\left(x\right)=\left(x^2+2x+10\right)^3\\</p>\\<p>R'\left(x\right)=3\left(x^2+2x+10\right)^2\left(2x+2\right)=\left(x^2+2x+10\right)^2\left(6x+6\right)\\</p>\\<p>\gcd\left(\left(x^2+2x+10\right)^3,\left(x^2+2x+10\right)^2\left(6x+6\right)\right)=\left(x^2+2x+10\right)^2</p>\\<p>

 

\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}}{\left(x^2+2x+10\right)^2}+\int{\frac{b_{1}x+b_{0}}{\left(x^2+2x+10\right)}}\\</p>\\<p>\frac{1}{\left(x^2+2x+10\right)^3}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+2x+10\right)^2-2\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(x^2+2x+10\right)\left(2x+2\right)}{\left(x^2+2x+10\right)^4}+\frac{b_{1}x+b_{0}}{\left(x^2+2x+10\right)}\\</p>\\<p>\frac{1}{\left(x^2+2x+10\right)^3}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+2x+10\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(4x+4\right)+\left(b_{1}x+b_{0}\right)\left(x^2+2x+10\right)^2}{\left(x^2+2x+10\right)^3}\\</p>\\<p>1=\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+2x+10\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(4x+4\right)+\left(b_{1}x+b_{0}\right)\left(\left(x+1\right)^2+9\right)^2\\</p>\\<p>1=\left(3a_{3}x^4+2a_{2}x^3+a_{1}x^2+6a_{3}x^3+4a_{2}x^2+2a_{1}x+30a_{3}x^2+20a_{2}x+10a_{1}\right)-\left(4a_{3}x^4+4a_{2}x^3+4a_{1}x^2+4a_{0}x+4a_{3}x^3+4a_{2}x^2+4a_{1}x+4a_{0}\right)+\left(b_{1}x+b_{0}\right)\left(\left(x+1\right)^4+18\left(x+1\right)^2+81\right)\\</p>\\<p>1=\left(3a_{3}x^4+2a_{2}x^3+a_{1}x^2+6a_{3}x^3+4a_{2}x^2+2a_{1}x+30a_{3}x^2+20a_{2}x+10a_{1}\right)-\left(4a_{3}x^4+4a_{2}x^3+4a_{1}x^2+4a_{0}x+4a_{3}x^3+4a_{2}x^2+4a_{1}x+4a_{0}\right)+\\</p>\\<p>\left(b_{1}x+b_{0}\right)\left(x^4+4x^3+6x^2+4x+1+18x^2+36x+18+81\right)\\</p>\\<p>1=\left(3a_{3}x^4+2a_{2}x^3+a_{1}x^2+6a_{3}x^3+4a_{2}x^2+2a_{1}x+30a_{3}x^2+20a_{2}x+10a_{1}\right)-\left(4a_{3}x^4+4a_{2}x^3+4a_{1}x^2+4a_{0}x+4a_{3}x^3+4a_{2}x^2+4a_{1}x+4a_{0}\right)+\\</p>\\<p>\left(b_{1}x+b_{0}\right)\left(x^4+4x^3+24x^2+40x+100\right)\\</p>\\<p>1=\left(3a_{3}x^4+2a_{2}x^3+a_{1}x^2+6a_{3}x^3+4a_{2}x^2+2a_{1}x+30a_{3}x^2+20a_{2}x+10a_{1}\right)-\left(4a_{3}x^4+4a_{2}x^3+4a_{1}x^2+4a_{0}x+4a_{3}x^3+4a_{2}x^2+4a_{1}x+4a_{0}\right)+\\</p>\\<p>\left(b_{1}x^5+4b_{1}x^4+24b_{1}x^3+40b_{1}x^2+100b_{1}x+b_{0}x^4+4b_{0}x^3+24b_{0}x^2+40b_{0}x+100b_{0}\right)\\</p>\\<p>1=b_{1}x^5+\left(4b_{1}+b_{0}-a_{3}\right)x^4+\left(24b_{1}+4b_{0}-2a_{2}+2a_{3}\right)x^3+\left(40b_{1}+24b_{0}+30a_{3}-3a_{1}\right)x^2+\left(100b_{1}+40b_{0}-2a_{1}-4a_{0}+20a_{2}\right)x+\left(100b_{0}+10a_{1}-4a_{0}\right)\\</p>\\<p>

 

</p>\\<p>\begin{cases}b_{1}=0\\4b_{1}+b_{0}-a_{3}=0\\24b_{1}+4b_{0}-2a_{2}+2a_{3}=0\\40b_{1}+24b_{0}+30a_{3}-3a_{1}=0\\100b_{1}+40b_{0}-2a_{1}-4a_{0}+20a_{2}=0\\100b_{0}+10a_{1}-4a_{0}=1\end{cases}\\</p>\\<p>\begin{cases}b_{1}=0\\b_{0}=a_{3}\\-a_{2}+3a_{3}=0\\18a_{3}-a_{1}=0\\40a_{3}-2a_{1}-4a_{0}+20a_{2}=0\\100a_{3}+10a_{1}-4a_{0}=1\end{cases}\\</p>\\<p>\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=3a_{3}=0\\a_{1}=18a_{3}\\16a_{3}=a_{0}\\216a_{3}=1\end{cases}\\</p>\\<p>\begin{cases}b_{1}=0\\216b_{0}=1\\216a_{2}=3\\216a_{1}=18\\216a_{0}=16\\216a_{3}=1\end{cases}\\<br>\\

 

\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{1}{216}\cdot\frac{x^3+3x^2+18x+16}{\left(x^2+2x+10\right)^2}+\frac{1}{216}\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)}}\\</p>\\<p>\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{1}{216}\cdot\frac{x^3+3x^2+18x+16}{\left(x^2+2x+10\right)^2}+\frac{1}{216}\int{\frac{\mbox{d}x}{\left(x+1\right)^2+9}}\\</p>\\<p>\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{1}{216}\cdot\frac{x^3+3x^2+18x+16}{\left(x^2+2x+10\right)^2}+\frac{1}{216}\cdot\frac{1}{9}\int{\frac{\mbox{d}x}{1+\left(\frac{x+1}{3}\right)^2}}\\</p>\\<p>\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{1}{216}\cdot\frac{x^3+3x^2+18x+16}{\left(x^2+2x+10\right)^2}+\frac{1}{216}\cdot\frac{1}{3}\int{\frac{\frac{1}{3}\mbox{d}x}{1+\left(\frac{x+1}{3}\right)^2}}\\</p>\\<p>\int{\frac{\mbox{d}x}{\left(x^2+2x+10\right)^3}}=\frac{1}{216}\cdot\frac{x^3+3x^2+18x+16}{\left(x^2+2x+10\right)^2}+\frac{1}{648}\arctan{\left(\frac{x+1}{3}\right)}+C</p>\\<p>


Użytkownik Jarekzulus edytował ten post 13.08.2015 - 13:58

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#24 Jarekzulus

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Napisano 13.08.2015 - 14:00

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Najwyższa ocena

i) Przez części w poście 10

 

\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}

 

R\left(x\right)=\left(x^2+4x+5\right)^2\left(x^2+4\right)^2\\</p>\\<p>R'\left(x\right)=2\left(x^2+4x+5\right)\left(2x+4\right)\left(x^2+4\right)^2+2\left(x^2+4x+5\right)^2\left(x^2+4\right)\cdot 2x=\left(x^2+4x+5\right)\left(x^2+4\right)\left(\left(4x+8\right)\left(x^2+4\right)+4x\left(x^2+4x+5\right)\right)\\</p>\\<p>R'\left(x\right)=\left(x^2+4x+5\right)\left(x^2+4\right)\left(4x^3+8x^2+16x+32+4x^3+16x^2+20x\right)=\left(x^2+4x+5\right)\left(x^2+4\right)\left(8x^3+24x^2+36x+32\right)\\</p>\\<p>\gcd\left(\left(x^2+4x+5\right)^2\left(x^2+4\right)^2,\left(x^2+4x+5\right)\left(x^2+4\right)\left(8x^3+24x^2+36x+32\right)\right)=\left(x^2+4x+5\right)\left(x^2+4\right)</p>\\<p>

 

\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=\frac{a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}}{\left(x^2+4x+5\right)\left(x^2+4\right)}+\int{\frac{b_{3}x^3+b_{2}x^2+b_{1}x+b_{0}}{\left(x^2+4x+5\right)\left(x^2+4\right)}\mbox{d}x}\\</p>\\<p>\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+4x+5\right)\left(x^2+4\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(\left(2x+4\right)\left(x^2+4\right)+2x\left(x^2+4x+5\right)\right)}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}+\frac{b_{3}x^3+b_{2}x^2+b_{1}x+b_{0}}{\left(x^2+4x+5\right)\left(x^2+4\right)}\\</p>\\<p>\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^4+4x^3+9x^2+16x+20\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(4x^3+12x^2+18x+16\right)+\left(b_{3}x^3+b_{2}x^2+b_{1}x+b_{0}\right)\left(x^4+4x^3+9x^2+16x+20\right)}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\\</p>\\<p>x^5-x^4-26x^2-24x-25=\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^4+4x^3+9x^2+16x+20\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(4x^3+12x^2+18x+16\right)+\left(b_{3}x^3+b_{2}x^2+b_{1}x+b_{0}\right)\left(x^4+4x^3+9x^2+16x+20\right)\\</p>\\<p>x^5-x^4-26x^2-24x-25=\left(3a_{3}x^6+12a_{3}x^5+27a_{3}x^4+48a_{3}x^3+60a_{3}x^2+2a_{2}x^5+8a_{2}x^4+18a_{2}x^3+32a_{2}x^2+40a_{2}x+a_{1}x^4+4a_{1}x^3+9a_{1}x^2+16a_{1}x+20a_{1}\right)-\\</p>\\<p>\left(4a_{3}x^6+12a_{3}x^5+18a_{3}x^4+16a_{3}x^3+4a_{2}x^5+12a_{2}x^4+18a_{2}x^3+16a_{2}x^2+4a_{1}x^4+12a_{1}x^3+18a_{1}x^2+16a_{1}x+4a_{0}x^3+12a_{0}x^2+18a_{0}x+16a_{0}\right)+\\</p>\\<p>\left(b_{3}x^7+4b_{3}x^6+9b_{3}x^5+16b_{3}x^4+20b_{3}x^3+b_{2}x^6+4b_{2}x^5+9b_{2}x^4+16b_{2}x^3+20b_{2}x^2+b_{1}x^5+4b_{1}x^4+9b_{1}x^3+16b_{1}x^2+20b_{1}x+b_{0}x^4+4b_{0}x^3+9b_{0}x^2+16b_{0}x+20b_{0}\right)\\</p>\\<p>x^5-x^4-26x^2-24x-25=b_{3}x^7+\left(4b_{3}+b_{2}-a_{3}\right)x^6+\left(9b_{3}+4b_{2}+b_{1}-2a_{2}\right)x^5+\left(16b_{3}+9b_{2}+4b_{1}+b_{0}+9a_{3}-4a_{2}-3a_{1}\right)x^4+\left(20b_{3}+16b_{2}+9b_{1}+4b_{0}+32a_{3}-8a_{1}-4a_{0}\right)x^3+\\</p>\\<p>\left(20b_{2}+16b_{1}+9b_{0}+60a_{3}+16a_{2}-9a_{1}-12a_{0}\right)x^2+\left(20b_{1}+16b_{0}+40a_{2}-18a_{0}\right)x+\left(20b_{0}+20a_{1}-16a_{0}\right)\\</p>\\<p>

 

 

</p>\\<p>\begin{cases}b_{3}=0\\4b_{3}+b_{2}-a_{3}=0\\9b_{3}+4b_{2}+b_{1}-2a_{2}=1\\16b_{3}+9b_{2}+4b_{1}+b_{0}+9a_{3}-4a_{2}-3a_{1}=-1\\20b_{3}+16b_{2}+9b_{1}+4b_{0}+32a_{3}-8a_{1}-4a_{0}=0\\20b_{2}+16b_{1}+9b_{0}+60a_{3}+16a_{2}-9a_{1}-12a_{0}=-26\\20b_{1}+16b_{0}+40a_{2}-18a_{0}=-24\\20b_{0}+20a_{1}-16a_{0}=-25\end{cases}\\</p>\\<p>\begin{cases}b_{3}=0\\8b_{2}=-9\\8b_{1}=-4\\8b_{0}=-37\\8a_{3}=-9\\8a_{2}=-24\\8a_{1}=-37\\8a_{0}=-80\end{cases}\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\int{\frac{9x^2+4x+37}{\left(x^2+4x+5\right)\left(x^2+4\right)}\mbox{d}x}\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\int{\frac{x^2+4x+5+8x^2+32}{\left(x^2+4x+5\right)\left(x^2+4\right)}\mbox{d}x}\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\int{\frac{\left(x^2+4x+5\right)+8\left(x^2+4\right)}{\left(x^2+4x+5\right)\left(x^2+4\right)}\mbox{d}x}\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\left(\int{\frac{\mbox{d}x}{x^2+4}}+8\int{\frac{\mbox{d}x}{x^2+4x+5}}\right)\\</p>\\<p></p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\left(\frac{1}{2}\int{\frac{\frac{1}{2}\mbox{d}x}{1+\left(\frac{x}{2}\right)^2}}+8\int{\frac{\mbox{d}x}{1+\left(x+2\right)^2}}\right)\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{8}\left(\frac{1}{2}\arctan{\left(\frac{x}{2}\right)}+8\arctan{\left(x+2\right)}\right)+C\\</p>\\<p>\int{\frac{x^5-x^4-26x^2-24x-25}{\left(x^2+4x+5\right)^2\left(x^2+4\right)^2}\mbox{d}x}=-\frac{1}{8}\cdot\frac{9x^3+24x^2+37x+80}{\left(x^2+4x+5\right)\left(x^2+4\right)}-\frac{1}{16}\arctan{\left(\frac{x}{2}\right)}-\arctan{\left(x+2\right)}+C<br>\\


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:wave: :wave: :wave: Jeśli rzuciłem choć promyczek światła na problem który postawiłeś - podziękuj. pre_1433974176__syg.jpgNad kreską


#25 Mariusz M

Mariusz M

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Napisano 13.08.2015 - 19:00

l)  Metodą na rozkład policzono w poście 13

 

\int{\frac{9}{5x^2\left(3-2x^2\right)^3}\mbox{d}x}

 

R\left(x\right)=x^2\left(3-2x^2\right)^3\\</p>\\<p>R'\left(x\right)=2x\left(3-2x^2\right)^3+3x^2\left(3-2x^2\right)^2\cdot\left(-4x\right)=x\left(3-2x^2\right)^2\left(6-4x^2-12x^2\right)\\</p>\\<p>R'\left(x\right)=x\left(3-2x^2\right)^2\left(6-16x^2\right)\\</p>\\<p>\gcd{\left(x^2\left(3-2x^2\right)^3,x\left(3-2x^2\right)^2\left(6-16x^2\right)\right)}=x\left(3-2x^2\right)^2</p>\\<p>

 

\int{\frac{9}{5x^2\left(3-2x^2\right)^3}\mbox{d}x}=\frac{a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}}{x\left(3-2x^2\right)^2}+\int{\frac{b_{2}x^2+b_{1}x+b_{0}}{x\left(3-2x^2\right)}\mbox{d}x}\\</p>\\<p>\frac{9}{5x^2\left(3-2x^2\right)^3}=\frac{\left(4a_{4}x^3+3a_{3}x^2+2a_{2}x+a_{1}\right)x\left(3-2x^2\right)^2-\left(a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(\left(3-2x^2\right)^2+x\left(3-2x^2\right)\left(-8x\right)\right)}{x^2\left(3-2x^2\right)^4}+\frac{b_{2}x^2+b_{1}x+b_{0}}{x\left(3-2x^2\right)}\\</p>\\<p>\frac{9}{5x^2\left(3-2x^2\right)^3}=\frac{\left(4a_{4}x^3+3a_{3}x^2+2a_{2}x+a_{1}\right)x\left(3-2x^2\right)^2-\left(a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(3-2x^2\right)\left(3-10x^2\right)}{x^2\left(3-2x^2\right)^4}+\frac{b_{2}x^2+b_{1}x+b_{0}}{x\left(3-2x^2\right)}\\</p>\\<p>\frac{9}{5x^2\left(3-2x^2\right)^3}=\frac{\left(4a_{4}x^3+3a_{3}x^2+2a_{2}x+a_{1}\right)\left(3x-2x^3\right)-\left(a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(3-10x^2\right)+\left(b_{2}x^2+b_{1}x+b_{0}\right)x\left(3-2x^2\right)^2}{x^2\left(3-2x^2\right)^3}\\</p>\\<p>\frac{9}{5}=\left(4a_{4}x^3+3a_{3}x^2+2a_{2}x+a_{1}\right)\left(3x-2x^3\right)-\left(a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(3-10x^2\right)+\left(b_{2}x^2+b_{1}x+b_{0}\right)\left(9x-12x^3+4x^5\right)\\</p>\\<p>\frac{9}{5}=\left(12a_{4}x^4+9a_{3}x^3+6a_{2}x^2+3a_{1}x-8a_{4}x^6-6a_{3}x^5-4a_{2}x^4-2a_{1}x^3\right)-\left(3a_{4}x^4+3a_{3}x^3+3a_{2}x^2+3a_{1}x+3a_{0}-10a_{4}x^6-10a_{3}x^5-10a_{2}x^4-10a_{1}x^3-10a_{0}x^2\right)\\+\left(9b_{2}x^3-12b_{2}x^5+4b_{2}x^7+9b_{1}x^2-12b_{1}x^4+4b_{1}x^6+9b_{0}x-12b_{0}x^3+4b_{0}x^5\right)\\</p>\\<p>\frac{9}{5}=4b_{2}x^7+\left(4b_{1}+2a_{4}\right)x^6+\left(-12b_{2}+4b_{0}+4a_{3}\right)x^5+\left(-12b_{1}+9a_{4}+6a_{2}\right)x^4+\left(9b_{2}-12b_{0}+6a_{3}+8a_{1}\right)x^3+\\</p>\\<p>\left(9b_{1}+3a_{2}+10a_{0}\right)x^2+9b_{0}x-3a_{0}\\</p>\\<p>

 

</p>\\<p>\begin{cases}4b_{2}=0\\4b_{1}+2a_{4}=0\\-12b_{2}+4b_{0}+4a_{3}=0\\-12b_{1}+9a_{4}+6a_{2}=0\\9b_{2}-12b_{0}+6a_{3}+8a_{1}=0\\9b_{1}+3a_{2}+10a_{0}=0\\9b_{0}=0\\-3a_{0}=\frac{9}{5}\end{cases}\\</p>\\<p>\begin{cases}b_{2}=0\\2b_{1}=-a_{4}\\a_{3}=0\\5a_{4}+2a_{2}=0\\a_{1}=0\\2a_{2}=4+3a_{4}\\b_{0}=0\\a_{0}=-\frac{3}{5}\end{cases}\\</p>\\<p>\begin{cases}b_{2}=0\\2b_{1}=-a_{4}\\a_{3}=0\\2a_{4}=-1\\a_{1}=0\\2a_{2}=4+3a_{4}\\b_{0}=0\\a_{0}=-\frac{3}{5}\end{cases}\\</p>\\<p>\begin{cases}b_{2}=0\\4b_{1}=1\\a_{3}=0\\2a_{4}=-1\\a_{1}=0\\4a_{2}=5\\b_{0}=0\\a_{0}=-\frac{3}{5}\end{cases}\\</p>\\<p>

 

\int{\frac{9}{5x^2\left(3-2x^2\right)^3}\mbox{d}x}=\frac{\frac{-1}{2}x^4+\frac{5}{4}x^2-\frac{3}{5}}{x\left(3-2x^2\right)^2}+\frac{1}{4}\int{\frac{\mbox{d}x}{3-2x^2}}\\</p>\\<p>\frac{1}{4}\int{\frac{\mbox{d}x}{\left(3-2x^2\right)}}=\int{\frac{p}{\sqrt{3}-\sqrt{2}x}\mbox{d}x}+\int{\frac{p}{\sqrt{3}-\sqrt{2}x}\mbox{d}x}\\</p>\\<p>\frac{1}{4}\cdot\frac{1}{3-2x^2}=\frac{p}{\sqrt{3}-\sqrt{2}x}+\frac{q}{\sqrt{3}+\sqrt{2}x}\\</p>\\<p>\frac{1}{4}=p\left(\sqrt{3}+\sqrt{2}x\right)+q\left(\sqrt{3}-\sqrt{2}x\right)\\</p>\\<p>\frac{1}{4}=\left(\sqrt{2}p-\sqrt{2}q\right)x+\left(\sqrt{3}p+\sqrt{3}q\right)\\</p>\\<p>\begin{cases}\sqrt{2}p-\sqrt{2}q=0\\\sqrt{3}p+\sqrt{3}q=\frac{1}{4}\end{cases}\\</p>\\<p>\begin{cases}q=p\\p=\frac{\sqrt{3}}{24}\end{cases}\\</p>\\<p>\int{\frac{9}{5x^2\left(3-2x^2\right)^3}\mbox{d}x}=\frac{\frac{-1}{2}x^4+\frac{5}{4}x^2-\frac{3}{5}}{x\left(3-2x^2\right)^2}+\frac{\sqrt{6}}{48}\left(\int{\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}x}\mbox{d}x}-\int{\frac{\left(-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}x}\mbox{d}x}\right)\\</p>\\<p>\int{\frac{9}{5x^2\left(3-2x^2\right)^3}\mbox{d}x}=\frac{\frac{-1}{2}x^4+\frac{5}{4}x^2-\frac{3}{5}}{x\left(3-2x^2\right)^2}+\frac{\sqrt{6}}{48}\ln{\left|\frac{\sqrt{3}+\sqrt{2}x}{\sqrt{3}-\sqrt{2}x}\right|}+C</p>\\<p>


Użytkownik Jarekzulus edytował ten post 13.08.2015 - 20:25

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