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Sa se calculeze[tex] \int\limits^1_0f( {x}) \, dx [/tex] f(x)=1/x+2

Răspuns :

[tex]\int_{0}^{1}\frac{1}{x+2}dx=ln(x+2)|_{0}^{1}=ln(1+2)-ln(0+2)=ln3-ln2=ln\frac{3}{2}[/tex]