<s>[INST] Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve 

$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$ 

through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration. 

Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$ 

Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
 [/INST] I don't know what is GMT. I looked it up on Wikipedia and it states $u = x - 1/x$ as Cauchy–Schlömilch substitution. Here is my solution that splits the integral into 4 integrals, each one of them uses $u = x - 1/x$. I will happily delete the answer if it is useless or wrong :)

$$\int^{\infty}_{-\infty} \dfrac{1}{1+ x^8} dx = -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 - \sqrt{2}}{x^4 - \sqrt 2 x^2 + 1} + \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + \sqrt{2}}{x^4 + \sqrt2 x^2 + 1} \\= -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} +\dfrac{\sqrt{2} +1}{2 \sqrt{2}} \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} \\+ \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1}+\dfrac{\sqrt2 -1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} $$

$$\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} = \int^{\infty}_{-\infty} \dfrac{1 + 1/x^2}{ (x -1/x)^2 + 2 - \sqrt 2}$$
$u = x - 1/x, du = 1 + 1/x^2$
$$\int^{\infty}_{-\infty} \dfrac{1}{u^2 + (2 - \sqrt 2 )} du $$

$$J =  \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} = 2 \int^{\infty}_{0} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1}$$
$x = 1/t, dx = -1/t^2 dt$
$$J = 2 \int^{\infty}_{0} \dfrac{t^2}{t^4 - \sqrt 2 t^2 + 1} dt$$
Adding the two equations, 
$$J = \int^{\infty}_{0} \dfrac{t^2 + 1}{t^4 - \sqrt 2 t^2 + 1} dt$$
This can be solved with $ u = t - 1/t$ like the previous equation.

$$\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1} = \int^{\infty}_{-\infty}\dfrac{1 + 1/x^2 }{(x - 1/x)^2 + \sqrt2 + 2}$$
Similar to first integral, now.

$$I = \int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} = 2\int^{\infty}_{0}\dfrac{1}{x^4 + \sqrt2 x^2 + 1}$$
$x = 1/t, dx = -1/t^2 dt$ 
$$ I = \int^{\infty}_{0}\dfrac{t^2 + 1}{t^4 + \sqrt2 t^2 + 1} dt $$
Can be solved with $u = t - 1/t$ like other integrals here.
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