Calculati
[tex]\int\limits^e_2 {\frac{1}{[(x-1)e^{-x}]e^x } } \, dx[/tex]

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Răspuns:

ln(e-1)

Explicație pas cu pas:

Prelucram raportul intalnit sub integrala:

[tex]\frac{1}{[(x-1)*e^{-x}]*e^x}=\frac{1}{(x-1)*e^{-x}*e^x}=\frac{1}{(x-1)*e^{x-x}}=\frac{1}{x-1}[/tex]

Integrala devine:

[tex]I=\int\limits^e_2 \frac{1}{x-1} \, dx =ln(x-1)|^e_2=ln(e-1)-ln(2-1)=ln(e-1)-ln1=ln(e-1)-0=ln(e-1)[/tex]