??

[sweetgirl87]

finn integralene.

[symbol:integral] [symbol:rot] x lnx dx

[symbol:integral] (x-1)^2 e^x dx

[Janhaa]

finn integralene.
[symbol:integral] [symbol:rot] x lnx dx

[tex]I\,=\,\int \sqrt{x} ln(x)\,dx[/tex]

bruk delvis integrasjon; v'= [symbol:rot] x , v=(2/3)x[sup]3/2[/sup] ,
u=ln(x) og u'= 1/x

[tex]I\,=\,{2\over 3}x^{3\over 2}ln(x)\,-\,{2\over 3}\int {1\over x}{x^{3\over 2}}dx[/tex]

[tex]I\,=\,{2\over 3}x^{3\over 2}ln(x)\,-\,{2\over 3}\int {x^{1\over 2}}dx\,=\,[/tex][tex]{2\over 3}x^{3\over 2}ln(x)\,-\,{2\over 3}({2\over 3}{x^{3\over 2}})\,+\,C[/tex]

[tex]I\,=\,{2\over 3}x^{3\over 2}ln(x)\,-\,[/tex][tex]({4\over 9}{x^{3\over 2}})\,+\,C\;=\;[/tex][tex]{2\over 3}x^{3\over 2}(ln(x)\,-\,{2\over 3})\,+\,C[/tex]

[sEirik]

[tex]J = \int (x-1)^2 e^x {\rm d}x[/tex]

To valgmuligheter:
1) Variabelskifte, u = x - 1. Så bruker du delvis integrasjon to ganger.
2) Bruk at [tex](x-1)^2 = x^2 - 2x + 1[/tex], delvis integrasjon to ganger.