Question

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

Answer 1

................................

Answer 2

Răspuns:

ln(e-1)

Explicație pas cu pas:

Prelucram raportul intalnit sub integrala:

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

Integrala devine:

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