Integral of $$$\frac{1}{x^{n} + 1}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{x^{n} + 1}\, dx$$$.

This integral does not have a closed form:

$${\color{red}{\int{\frac{1}{x^{n} + 1} d x}}} = {\color{red}{x {{}_{2}F_{1}\left(\begin{matrix} 1, \frac{1}{n} \\ 1 + \frac{1}{n} \end{matrix}\middle| {- x^{n}} \right)}}}$$

Therefore,

$$\int{\frac{1}{x^{n} + 1} d x} = x {{}_{2}F_{1}\left(\begin{matrix} 1, \frac{1}{n} \\ 1 + \frac{1}{n} \end{matrix}\middle| {- x^{n}} \right)}$$

Simplify:

$$\int{\frac{1}{x^{n} + 1} d x} = \frac{x \Phi\left(x^{n} e^{i \pi}, 1, \frac{1}{n}\right)}{n}$$

Add the constant of integration:

$$\int{\frac{1}{x^{n} + 1} d x} = \frac{x \Phi\left(x^{n} e^{i \pi}, 1, \frac{1}{n}\right)}{n}+C$$

$$$\int \frac{1}{x^{n} + 1}\, dx = \frac{x \Phi\left(x^{n} e^{i \pi}, 1, \frac{1}{n}\right)}{n} + C$$$A