(ax + b)ndx =

(n -1)

(ax + b)-1dx =

(1/a)· ln |ax + b|

x· (ax + b)ndx =

(n -1 et n -2)

x· (ax + b)-1dx = x/a - (b/a2)· ln |ax + b|

x· (ax + b)-2dx =

+ (1/a2)· ln |ax + b|

dx =