Close-Form General Solution:

Question

Is there a general solution to the integral $$\int \frac{1}{x^n+1}dx$$ where $x\in ℝ$ and $n\in \mathbb{Z}$?

Answer

$$\begin{array}{rl}\int \frac{dx}{x^n+1}& =\int (1+x^n{)}^{-1}dx\\ & =\int (1-x^n+x^{2n}-x^{3n}+x^{4n}-x^{5n}...)dx\\ & =x(1-\frac{x^n}{n+1}+\frac{x^{2n}}{2n+1}-\frac{x^{3n}}{3n+1}+\frac{x^{4n}}{4n+1}-\frac{x^{5n}}{5n+1}...)+C\\ & =x(1+\frac{1.\frac{1}{n}}{1+\frac{1}{n}}\frac{(-x^n{)}^1}{1!}+\frac{1.2.(\frac{1}{n}(\frac{1}{n}+1))}{(\frac{1}{n}+1)(\frac{1}{n}+2)}\frac{(-x^n{)}^2}{2!}+)+C\\ & =x_2{F}_1(1,\frac{1}{n};1+\frac{1}{n};-x^n)+C\end{array}$$ Here, we have used Gaussian Hypergeometric function as I believed $b$ and $c$ have telescopic ratio:) and $(a{)}_k$ is simply $k!$ or else considering a generalized form $$I=\int \frac{x^{l-1}}{x^n+1}dx$$ if $n$ is even: $$I=-\frac{1}{n}\sum _{r=1}^{\frac{n}{2}}\cos \left(\frac{(2r-1)l\pi }{n}\right)\log (x^2-2x\cos \left(\frac{(2r-1)\pi }{n}\right)+1)+\frac{2}{n}\sum _{r=1}^{\frac{n}{2}}\sin \left(\frac{(2r-1)l\pi }{n}\right){\tan }^{-1}\left(\frac{x-\cos ((2r-1)\pi /n)}{\sin ((2r-1)\pi /n)}\right)$$ if $n$ is odd: $$I=\frac{(-1{)}^{l-1}}{n}\log (x+1)-\frac{1}{n}\sum _{r=1}^{\frac{n-1}{2}}\cos \left(\frac{(2r-1)l\pi }{n}\right)\log (x^2-2x\cos \left(\frac{(2r-1)\pi }{n}\right)+1)+\frac{2}{n}\sum _{r=1}^{\frac{n-1}{2}}\sin \left(\frac{(2r-1)l\pi }{n}\right){\tan }^{-1}\left(\frac{x-\cos ((2r-1)\pi /n)}{\sin ((2r-1)\pi /n)}\right)$$