d f(x)=[xctg(x+5)-ln5x] `dx=
[(xctg(x+5)) `-(ln5x) `]*dx=
[ (x ` )*ctg (x+5)+x*(ctg(x+5) `-(ln5x) `] dx relatia 1
x `=1
x+5=u =>( ctgu) `=-u `/sin²u=- 1/sin²(x+5)
(ln5x) ` 5x=u (lnu)`= u `/u= 5/5x=1/x
Introduci aceste valori in relatia 1 si obtii
df(x)=[ctg(x+5)-x/sin²(x+5)-1/x)]dx
Intrebari?