Integral de $$$\frac{1}{x^{n} + 1}$$$ con respecto a $$$x$$$

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

Esta integral no tiene una forma cerrada:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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