[tex]I_{n+1}=\int_1^e{x\ln^{n+1}xdx}=\int_1^e{(\frac{x^2}{2})'\ln^{n+1}xdx}=\\
\frac{x^2}{2}\ln^{n+1}x|_1^e-(n+1)\int_1^e{\frac{x^2}{2}\ln^{n}x*\frac{1}{x}dx}=\\
\frac{x^2}{2}\ln^{n+1}x|_1^e-\frac{n+1}{2}\int_1^e{x\ln^{n}xdx}=e^2-\frac{n+1}{2}I_n[/tex]
Am folosit metoda de integrare prin parti.
Daca sunt neclaritati poti sa ma mai intrebi.
