It gets complicated pretty quickly.
For $n = 1$ $$ \int_{-\infty}^{\infty} \frac{1}{x^2+1}dx = \arctan(x)_{-\infty}^{\infty}$$ For $n=2$ $$\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx = \left[\frac{1}{4\sqrt{2}}\log{\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}} + \frac{1}{2\sqrt2}\arctan{\frac{\sqrt2x}{1-x^2}}\right]_{-\infty}^{\infty}$$ Is there a formula for all $n$ and if so where can I find it.
Use the known result
$$\int_0^\infty \frac {x^{a-1}}{x^m +1} = \frac\pi m \csc\frac{\pi a}m $$
Well, we have the following integral:
$$\mathscr{S}_\text{n}:=\int_\mathbb{R}\frac{1}{1+x^{2\text{n}}}\space\text{d}x\tag1$$
Where $\text{n}\in\mathbb{N}$.
Let's first observe that the integrand is an even function, this implies:
$$\mathscr{S}_\text{n}=2\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x\tag2$$
Now, we can split the integral:
$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\int_0^1\frac{1}{1+x^{2n}}\space\text{d}x+\int_0^1\frac{x^{2\text{n}-2}}{1+x^{2\text{n}}}\space\text{d}x\tag3$$
And it is not very hard to see that we can use an infinite series to rewrite the integrands:
$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\int_0^1\left(1-x^{2\text{n}}+x^{4\text{n}}-x^{6\text{n}}+\dots\right)\space\text{d}x+$$ $$\int_0^1\left(x^{2\text{n}-2}-x^{4\text{n}-2}+x^{6\text{n}-2}+\dots\right)\space\text{d}x\tag4$$
I will let you prove the fact:
$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\frac{1}{2\text{n}}\sum_{\text{k}\space\in\space\mathbb{Z}}\frac{(-1)^\text{k}}{\text{k}+\frac{1}{2\text{n}}}=\frac{\pi}{2\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)\tag5$$
The last step uses the result from "An Infinite Alternating Harmonic Series".
So, in the end we end up with:
$$\mathscr{S}_\text{n}:=\int_\mathbb{R}\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\frac{2\pi}{2\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)=\frac{\pi}{\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)$$
