[symbol:integral] x^2+3x+4 / x+1 altså de tre leddene delt på x+1.
Har kommet fra til at det beste er å skrive det slik:
x^2/x+1 + 3x/x+1 + 4/x+1
kan noen hjelpe meg?
Finn det ubest. integralet
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]\int \frac{x^2+3x+4}{x+1}\rm{d}x[/tex]
[tex]\int \frac{x^2}{x+1}\rm{d}x + \int \frac{3x}{x+1}\rm{d}x + \int \frac{4}{x+1}\rm{d}x[/tex]
[tex]I = I_1 + I_2 + I_3[/tex]
[tex]I_1 = \int \frac{x^2}{x+1}\rm{d}x[/tex]
Polynomdivisjon: [tex]x^2 \ : \ (x+1) = x-1 + \frac{1}{x+1}[/tex]
[tex]I_1 = \int (x-1 + \frac{1}{x+1})\rm{d}x = \frac{1}{2}x^2-x+\ln{|x+1|} + C[/tex]
[tex]I_2 = \int\frac{3x}{x+1}\rm{d}x[/tex]
[tex]u = x+1 \ , \ \rm{d}u = \rm{d}x[/tex]
[tex]I_2 = \int \frac{3u-3}{u}\rm{d}u = 3\int 1 - \frac{1}{u}\rm{d}u = 3(u-\ln{|u|} + C = 3x+3-3\ln{|x+1|} + C = 3x-3\ln{|x+1|} + C[/tex]
[tex]I_3 = 4\int\frac{1}{x+1}\rm{d}x = 4\ln{|x+1|} + C[/tex]
[tex]I = \frac{1}{2}x^2-x+\ln{|x+1|} + 3x-3\ln{|x+1|} + 4\ln{|x+1|} + C[/tex]
[tex]I = \frac{1}{2}x^2+2x + \ln{|\frac{(x+1)^5}{(x+1)^3}|} + C[/tex]
[tex]I = \frac{1}{2}x^2 + 2x + 2\ln{|x+1|} + C = \underline{\underline{\frac{1}{2}(x+2)^2 + 2\ln{|x+1|} + C}}[/tex]
[tex]\int \frac{x^2}{x+1}\rm{d}x + \int \frac{3x}{x+1}\rm{d}x + \int \frac{4}{x+1}\rm{d}x[/tex]
[tex]I = I_1 + I_2 + I_3[/tex]
[tex]I_1 = \int \frac{x^2}{x+1}\rm{d}x[/tex]
Polynomdivisjon: [tex]x^2 \ : \ (x+1) = x-1 + \frac{1}{x+1}[/tex]
[tex]I_1 = \int (x-1 + \frac{1}{x+1})\rm{d}x = \frac{1}{2}x^2-x+\ln{|x+1|} + C[/tex]
[tex]I_2 = \int\frac{3x}{x+1}\rm{d}x[/tex]
[tex]u = x+1 \ , \ \rm{d}u = \rm{d}x[/tex]
[tex]I_2 = \int \frac{3u-3}{u}\rm{d}u = 3\int 1 - \frac{1}{u}\rm{d}u = 3(u-\ln{|u|} + C = 3x+3-3\ln{|x+1|} + C = 3x-3\ln{|x+1|} + C[/tex]
[tex]I_3 = 4\int\frac{1}{x+1}\rm{d}x = 4\ln{|x+1|} + C[/tex]
[tex]I = \frac{1}{2}x^2-x+\ln{|x+1|} + 3x-3\ln{|x+1|} + 4\ln{|x+1|} + C[/tex]
[tex]I = \frac{1}{2}x^2+2x + \ln{|\frac{(x+1)^5}{(x+1)^3}|} + C[/tex]
[tex]I = \frac{1}{2}x^2 + 2x + 2\ln{|x+1|} + C = \underline{\underline{\frac{1}{2}(x+2)^2 + 2\ln{|x+1|} + C}}[/tex]
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- Registrert: 10/04-2008 17:58
Tusen takk!
Polynomdivisjon er en fin ting![Wink :wink:](./images/smilies/icon_wink.gif)
Polynomdivisjon er en fin ting
![Wink :wink:](./images/smilies/icon_wink.gif)