$$\int\frac1{ \sqrt{x^n + 1}}dx$$
This question popped up in my mind after we took trigonometric substitution in calc II. I tried solving it until I got $${\sinh}^{{\frac{2}{{n}}-{1}}}{\left(\theta\right)}$$ and then it got ugly afterwards.
I came up with some reduction formula, but then the values did not really work out when I tested them out with constants and limits.
Moreover, a quick search on the internet showed me that such type of questions are solved with complex analysis, Mobius transformation etc.
I think this integral is solvable for all infinite integras and finite rationals, but again I am not sure.
The antiderivative for general $n \neq 0$ is expressible in terms of a hypergeometric function or, just as well, the incomplete beta function: $$\color{#df0000}{\boxed{\int \frac{dx}{\sqrt{x^n + 1}} = x \cdot{}_2 F_1\left(\frac{1}{2}, \frac{1}{n}; 1 + \frac{1}{n}; -x^n\right) + C}} .$$ For most values of $n$ this expression cannot be written as a closed expression in elementary functions, but for certain special rational values it can. We can read off from the result of your hyperbolic trigonometric solution that there is a closed form for $n = \frac{1}{m}$ and $n = \frac{2}{m}$, $m \in \Bbb Z \setminus \{ 0 \}$.
If $n$ is nonzero and rational, say, $n = \frac{a}{b}$ for $a, b \in \Bbb Z \setminus \{0\}$. Then, the substitution $$x = u^b, \qquad dx = b u^{b - 1} \,du$$ transforms the integral to $$b \int \frac{u^{b - 1} \,du}{\sqrt{u^a + 1}} .$$
Of course, if $n = 0$, the integrand is constant.