Noen som kan derivere denne for meg?
[symbol:integral] x+1/ x^2 - 4x + 4
Integral
Moderators: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
Jeg regner med du mener integrere, og ikke derivere 
[tex] \int \frac{x+1}{x^2-4x+4} \ dx =\int \frac{x+1}{(x-2)^2} \ dx[/tex]
La [tex] u = x-2[/tex], [tex] \frac{du}{dx} = 1[/tex]
[tex] \int \frac{x+1}{x^2-4x+4} \ dx = \int \frac{u+3}{u^2} \ du = \ln (u) - \frac{3}{u} + C = \ln (x-2) - \frac{3}{x-2} + C [/tex]
[tex] \int \frac{x+1}{x^2-4x+4} \ dx =\int \frac{x+1}{(x-2)^2} \ dx[/tex]
La [tex] u = x-2[/tex], [tex] \frac{du}{dx} = 1[/tex]
[tex] \int \frac{x+1}{x^2-4x+4} \ dx = \int \frac{u+3}{u^2} \ du = \ln (u) - \frac{3}{u} + C = \ln (x-2) - \frac{3}{x-2} + C [/tex]