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Razvanantonio
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f:(-1,∞)→R , f(x)=ln(x+1)
Aratati ca functia F:(-1,∞)--->R , F(x)=(x+1)ln(x+1)-x este o primitiva a functiei f.

Răspuns :

[tex]\displaystyle F'(x)=((x+1)ln(x+1)-x=((x+1)ln(x+1))'-x'=\\ \\ =(x+1)'ln(x+1)+(x+1)(ln(x+1))'-1=\\ \\ =1 \cdot ln(x+1) +(x+1) \cdot \frac{(x+1)'}{(x+1)} -1=ln(x+1)+(x+1) \cdot \frac{1}{x+1} -1=\\ \\ =ln(x+1)+1-1=ln(x+1)=f(x)[/tex]