Răspuns:
Se face schimbarea de variabilă
[tex]x=\displaystyle\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt\\x=\frac{1}{2}\Rightarrow t=2, \ x=2\Rightarrow t=\frac{1}{2}[/tex]
[tex]I=\displaystyle\int_2^{\frac{1}{2}}\frac{-\ln t}{\frac{1}{t^2}+\frac{1}{t}+1}\cdot\left(-\frac{1}{t^2}\right)dt=\int_2^{\frac{1}{2}}\frac{t^2\cdot\ln t}{t^2+t+1}\cdot\frac{1}{t^2}dt=\\=\int_2^{\frac{1}{2}}\frac{\ln t}{t^2+t+1}dt=-\int_{\frac{1}{2}}^2\frac{\ln t}{t^2+t+1}=-I[/tex]
Rezultă
[tex]I=-I\Rightarrow 2I=0\Rightarrow I=0[/tex]
Explicație pas cu pas:
