\displaystyle{Am\ scris\ fractia\ ca\ x-\dfrac{x}{x^n+1},\ prima\ parte\ duce\ la\ \dfrac{9}2,\ iar\ cu\ schimbarea\ de\ variabila:}\\x^n=\dfrac{1}t\ pentru\int_0^3\dfrac{x}{x^n+1}dx\ldots\ \ldots\ \ldots\int_{1/3^n}^{+\infty}\dfrac{dt}{1+t}=ln(1+t)\Big|^{+\infty}_{1/3^n}.$
De aici m-am cam blocat
. O altă idee, poate ? Mulțumesc.Green eyes.