[tex]\displaystyle \lim_{n \to \infty} \int\limits^3_0 { \frac{x^{n+1}}{x^{n}+1} } \, dx =\\
= \lim_{n \to \infty} \int\limits^3_0 (x- \frac{x}{x^n+1})dx=\\
= \lim_{n \to \infty} ( \frac{x^2}{2}|_0^3- \int\limits^3_0 \frac{x}{x^n+1})dx=\\
= \lim_{n \to \infty} ( \frac{9}{2} - \int\limits^3_0 \frac{x}{x^n+1})dx= \frac{9}{2} - \frac{1}{2} =4\\
Obs. \lim_{n \to \infty} \int \limits^3_0 \frac{x}{x^n+1}dx=\lim_{n \to \infty} \int\limits^1_0 \frac{x}{x^n+1}dx+\lim_{n \to \infty} \int\limits^3_1 \frac{x}{x^n+1}dx[/tex]
[tex]\displaystyle \lim_{n \to \infty} \int\limits^1_0 \frac{x}{x^n+1}dx+\lim_{n \to \infty} \int\limits^3_1 \frac{x}{x^n+1}dx=\\
=\int\limits^1_0\lim_{n \to \infty} \frac{x}{x^n+1}dx+\int\limits^3_1 \lim_{n \to \infty} \frac{x}{x^n+1}dx=\\
=\int\limits^1_0\lim_{n \to \infty} \frac{x}{0+1}dx+\int\limits^3_1 0dx=\\
= \frac{x^2}{2} |_0^1= \frac{1}{2} [/tex]
