Integrate $\int_{0}^{\infty} \frac{dx}{1 + x^n}$ using the Residue Theorem I'm trying to compute the following integral:

\begin{align}
\int_{0}^{\infty} \frac{dx}{1 + x^n}, \quad n \geq 2.
\end{align}

Consider the function $f(z) = \frac{1}{1 + z^n}$. Consider the following contour, $\Gamma_n$:

We then have,
\begin{align}
\int_{\Gamma_n} = \int_{0}^{R} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{1}} \frac{dz}{1 + z^n} + \int_{\Gamma_n^{2}} \frac{dz}{1 + z^n},
\end{align}
where $\Gamma_n^{1}$ denotes an arc of the circle, $R e^{i \theta}, \theta \in [0, 2\pi] $. I have shown that the integral along this arc goes to zero as $R$ goes to $\infty$. I'm having trouble with the integral along the arc $\Gamma_n^{2}$. Here's what I have done:
Parametrize $\Gamma_n^{2}$ as,
\begin{align}
z(t) = \Big (1 - \frac{t}{R} \Big ) R e^{\frac{2 \pi i}{n}}, \quad t \in [0, R].
\end{align}
We then have,
\begin{align}
dz(t) & = - e^{\frac{2 \pi i}{n}} dt, \\
1 + z(t)^n & = 1 + R^n \Big (1 - \frac{t}{R} \Big )^n = 1 + R^n \sum_{k = 0}^{n} (-1)^k \Big ( \frac{t}{R} \Big )^k \\ 
& = 1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n.
\end{align}
Hence, we have that,
\begin{align}
\int_{\Gamma_n^{2}} \frac{dz}{1 + z^n} = - e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dt}{1 + R^n - t R^{n-1} + t^2 R^{n-2} - \cdots + (-1)^n t^n}.
\end{align}
I'm not sure how to proceed from here. The answer given states that this integral converges to,
\begin{align}
- e^{\frac{2 \pi i}{n}} \int_{0}^{\infty} \frac{dx}{1 + x^n}.
\end{align}

Note that $\Gamma_n^1$ is an arc of a circle, not a semicircle (in general).
Besides,\begin{align}\int_{\Gamma_n^2}\frac{\mathrm dz}{1+z^n}&=\int_0^R\frac{-e^{\frac{2\pi i}n}}{1+\left(1-\frac tR\right)^nR^n}\,\mathrm dt\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+(R-t)^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dt}{1+t^n}\\&=-e^{\frac{2\pi i}n}\int_0^R\frac{\mathrm dz}{1+z^n}.\end{align}Can you take it from here?
