$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:
\begin{equation}
 I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx
\end{equation}
I believe what I've done is correct, but I'm concerned that I may missed something (in particular whether it holds for all $r \neq 0$). So, here I have two questions (1) Is what I've done correct? and (2) What other methods can be employed that doesn't rely on complex analysis?
Here is the method I took:
First make the substitution $u = x^{\frac{1}{r}}$ to arrive at
\begin{equation}
 I = \frac{1}{n} \int_{0}^{\infty} \frac{1}{1 + u} \cdot u^{1 -\frac{1}{r}}\:du
\end{equation}
We now substitute $t = \frac{1}{1 + u}$ to arrive at:
\begin{align}
 I &= \frac{1}{r} \int_{1}^{0} t \cdot \left(\frac{1 - t}{t}\right)^{\frac{1}{r} -1}\frac{1}{t^2}\:dt = \frac{1}{r}\int_{0}^{1}t^{-\frac{1}{r}}\left(1 - t\right)^{ \frac{1}{r} - 1}\:dt \\
&= \frac{1}{r}B\left(1 - \frac{1}{n}, 1 + \frac{1}{r} - 1\right) = \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right) \\
&= \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right)
\end{align}
Wheer $B(a,b)$ is the Beta function. Using the relationship between the Beta and Gamma function we arrive at:
\begin{equation}
I = \frac{1}{r} \frac{\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)}{\Gamma\left(\frac{r - 1}{r} + \frac{1}{r}\right)} = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
And so, we arrive at:
\begin{equation}
 I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
for $r > 1$
As per KemonoChen's comment and others, we can employ Euler's Reflection Formula to position this result for $\frac{1}{r} \not \in \mathbb{Z}$ Here, as $r \in \mathbb{R}, r > 1 \rightarrow \frac{1}{r} \not \in \mathbb{Z}$ and so our formula holds.
\begin{equation}
 I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) = \frac{\pi}{r\sin\left(\frac{\pi}{r} \right)}
\end{equation}
Thank you also to Winther, Jjagmath, and MrTaurho's for their comments and corrections/clarifications.

Once again I will offer up a method that first converts the integral to a double integral. 
For $r > 0$, we begin by enforcing a substitution of $x \mapsto x^{1/r}$. Doing so yields
$$I = \frac{1}{r} \int_0^\infty \frac{x^{1/r - 1}}{1 + x} \, dx.$$
Now noting that
$$\frac{1}{1 + x} = \int_0^\infty e^{-u(1 + x)} \, du,$$
our integral can be rewritten as
$$I = \frac{1}{r} \int_0^\infty x^{1/r - 1} \int_0^\infty e^{-u (1 + x)} \, du \, dx,$$
or
$$I = \frac{1}{r} \int_0^\infty e^{-u} \int_0^\infty x^{1/r - 1} e^{-ux} \, dx \, du,$$
after changing the order of integration.
Next we enforce a substitution of $x \mapsto x/u$. This gives
\begin{align}
I &= \frac{1}{r} \int_0^\infty u^{- 1/r} e^{-u} \, du \int_0^\infty x^{1/r - 1} e^{-x} \, dx\\
&= \frac{1}{r} \Gamma \left (1 - \frac{1}{r} \right ) \Gamma \left (\frac{1}{r} \right )\\
&= \frac{\pi}{r \sin \left (\frac{\pi}{r} \right )},
\end{align}
where in the last line we have made use of Euler's reflexion formula for the gamma function. 
