Integral
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Delvis integrasjon:pevik skrev:[symbol:integral] 2x lnx dx
[tex]u^{\prime} = 2x \ \ u = \frac12 x^2[/tex]
[tex]v = \ln x \ \ v^{\prime} = \frac{1}{x}[/tex]
[tex]\int 2x \cdot \ln x dx = \frac12 x^2 \cdot \ln x - \int \frac12 x^2 \cdot \frac{1}{x} dx = \frac12 x^2 \cdot \ln x - \frac12 \int x dx = \underline{\underline{\frac12 x^2 \cdot \ln x - \frac14 x^2 + C}}[/tex]
Det der blir vel feil.
[tex]u^, = 2x \ , \ u = x^2 \\ v = \ln{x} \ , \ v^, = \frac{1}{x}[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2\ln{x} - \int x^2 \ \cdot \ \frac{1}{x}\rm{d}x[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2\ln{x} - \int x\rm{d}x[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2 \ln{x} - \frac{1}{2}x^2 + C = \underline{\underline{x^2(\ln{x} - \frac{1}{2}) + C}}[/tex]
[tex]u^, = 2x \ , \ u = x^2 \\ v = \ln{x} \ , \ v^, = \frac{1}{x}[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2\ln{x} - \int x^2 \ \cdot \ \frac{1}{x}\rm{d}x[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2\ln{x} - \int x\rm{d}x[/tex]
[tex]\int 2x\ln{x}\rm{d}x = x^2 \ln{x} - \frac{1}{2}x^2 + C = \underline{\underline{x^2(\ln{x} - \frac{1}{2}) + C}}[/tex]