Question

Am nevoie de ajutor: Se considera f, F: [1, ∞) → R, f(x)= ㏑x + $\frac{1}{x}$ si F(x)= (x+1)㏑x-x+1.
Sa se arate ca F este o primitiva a functiei f, care se anuleaza in x=1.

Answer 1

=================================

Answer 2

Derivezi F si     arati    ca     obtii  f

F `(x)=[(x+1)*lnx-x+1] `=(x+1) `*lnx+(x+1)*(lnx)`-x `+1 `=

lnx+(x+1)/x-1+0=

lnx+x/x+1/.x-1=

lnx+1+1/x-1=lnx+1/x

F(1)=(1+1)*ln1-1+1=

2*0+0=0