$$\lim_{n\to\infty}\int_0^{\infty}{\frac{x}{1+x^{n}}\,\mathrm dx}$$Consider the integral$$\int_{0}^{\infty} \frac{x}{1+x^n} \,\mathrm dx$$
This is an improper integral, but we can solve it using substitution or integration by parts.
Let $u = x^n$, then $\mathrm du = nx^{n-1}\,\mathrm dx$.
$$\int_{0}^{\infty} \frac{x}{1+x^n} \,\mathrm dx = \frac{1}{n} \int_{0}^{\infty} \frac{1}{1+u} \, \mathrm du$$
Now, integrate:
$$\frac{1}{n} \int_{0}^{\infty} \frac{1}{1+u} \,\mathrm du = \left.\frac{1}{n} \ln(1+u) \right|_{0}^{\infty}$$
Apply the limit as $n\to\infty$, the integral approaches:
$$\lim_{n\to\infty}\left. \frac{1}{n} \ln(1+u) \right|_{0}^{\infty}=\lim_{n\to\infty} \frac{\ln(1+\infty)-\ln(1)}{n}$$
Simplify. Since $\ln(1+\infty)=\infty$, we have:
$$\lim_{n\to\infty} \frac{\infty}{n}=\lim_{n\to\infty} \frac{1}{n} \cdot \infty=0$$
The final answer is: $\boxed{0}$. Is my solution correct?