$$\int_0^\infty \frac{1}{1+x^n} \, dx=\frac 1n \int_0^\infty \frac{x^{\frac 1n-1}}{1+x} \, dx=\frac 1n \left(\frac{\pi}{\sin \frac{\pi}n} \right)=\frac{\pi}n\csc\left(\frac{\pi}n\right)$$
Which rule or law has been applied after the first equality sign?
The "rule" that has been applied is
$$ \int_0^\infty \frac{x^{a - 1}}{1 + x} = \frac{\pi}{\sin \pi a}, \quad 0 < a < 1$$
If you don't know residue theory, one way to show this is to prove that the function
$$F(a) = \int_0^\infty \frac{x^{a-1}}{1 + x}\, dx,\quad 0 < a < 1$$
satisfies the differential equation
$$\ddot{F}F - \dot{F}^2 - F^4 = 0,\quad F(1/2) = \pi, \quad \dot{F}(1/2) = 0$$
and verify that the unique solution is $\pi/\sin \pi a$.
The substitution $$ z = x^n \\ dz = n x^{n-1} dx = n (z/x) \, dx \Rightarrow dx = \frac{1}{n} \frac{x}{z} dz = \frac{1}{n} \frac{z^{1/n}}{z} dz = \frac{1}{n} z^{(1/n)-1} dz $$ which leads to $$ \int\limits_0^\infty \frac{dx}{1+x^n} = \frac{1}{n}\int\limits_0^\infty \frac{z^{(1/n)-1}}{1+z} dz $$ was used and then they renamed $z$ back into $x$.
Letting $u=x^n$, $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{1+x^n} &=\frac1n\int_0^\infty\frac{u^{\frac1n-1}}{1+u}\,\mathrm{d}u\\[8pt] &=\tfrac1n\mathrm{B}\left(\tfrac1n,1-\tfrac1n\right)\\[4pt] &=\frac1n\frac{\Gamma\left(\frac1n\right)\Gamma\left(1-\frac1n\right)}{\Gamma(1)} \end{align} $$ where $\mathrm{B}(x,y)$ is the Beta Function and $\Gamma(x)$ is the Gamma Function.
We can use Euler's Reflection Formula, which says $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x) $$ A number of methods for evaluating this integral and a proof of the Euler's Reflection Formula can be found in this answer.
