Răspuns:a
Explicație pas cu pas:
a) Seaplica proprietatea F `(x)=f(x)
[(x+1)ln(x+1)-x] `=(x+1) `ln(x+1)+(x+1)*ln(x+!) `-x`=
ln(x+1)+(x+1)/(x+1)-1=ln(x+1)+1=1=
ln(x+1)
b)G(x)=(x+1)ln(x+1)-x+c
G(0)=3=>
(0+1)ln(0+1)-0+c=3
1*ln1+c=3
0+c=3
c=3
c)A=∫f(x)dx=F(x)║₀ᵃ⁻¹=F(e⁻¹)-F(0)=
(e⁻¹+1)*ln(e⁻¹+1)-e⁻⁻¹-[(0+1)ln(0+1)-0]=
e⁻¹ln(e⁻¹+1)-e⁻⁻¹=
e⁻⁻¹(ln(e⁻⁻¹+1)-1)