1. **Problem Statement:** Find the value of the integral $$\int_0^\infty \frac{1}{1+x^n} \, dx$$ for $$n > 1$$. 2. **Formula and Important Rules:** This integral is a known form related to the Beta and Gamma functions. The integral converges for $$n > 1$$. 3. **Using substitution and Beta function:** Recall the Beta function definition: $$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ 4. **Rewrite the integral:** Let $$x^n = t$$, so $$x = t^{1/n}$$ and $$dx = \frac{1}{n} t^{\frac{1}{n}-1} dt$$. 5. **Change limits:** When $$x=0$$, $$t=0$$; when $$x=\infty$$, $$t=\infty$$. 6. **Substitute into integral:** $$\int_0^\infty \frac{1}{1+x^n} dx = \int_0^\infty \frac{1}{1+t} \cdot \frac{1}{n} t^{\frac{1}{n}-1} dt = \frac{1}{n} \int_0^\infty \frac{t^{\frac{1}{n}-1}}{1+t} dt$$ 7. **Use Beta function integral form:** For $$0 < a < 1$$, $$\int_0^\infty \frac{t^{a-1}}{1+t} dt = \pi / \sin(\pi a)$$. Here, $$a = \frac{1}{n}$$. 8. **Final evaluation:** $$\int_0^\infty \frac{1}{1+x^n} dx = \frac{1}{n} \cdot \frac{\pi}{\sin\left(\frac{\pi}{n}\right)}$$ **Answer:** $$\boxed{\int_0^\infty \frac{1}{1+x^n} dx = \frac{\pi}{n \sin\left(\frac{\pi}{n}\right)}}$$