Seeking Methods to solve $\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}\:dx $

I can offer a shorter way to obtain the final result in terms of the Incomplete Gamma Function. Therby consider the following representation of the Incomplete Gamma Function which can be found here on MSE

$$ \Gamma(a,x)=\frac{e^{-x}x^{a}}{\Gamma(1-a)} \int_0^\infty \frac{e^{-t} t^{-a}}{x+t} dt\tag1$$

Getting back to your original integral and applying the substitution $x^n=t$ yields to the following

$$\begin{align} I_n=\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}dx&=\int_{0}^{\infty} \frac{e^{-t}}{1+t}\frac1nt^{1/n-1}dt\\ &=\frac1n\int_0^{\infty}\frac{e^{-t}t^{-(1-1/n)}}{1+t}dt \end{align}$$

The latter integral is in the form of $(1)$ with $a=1-1/n$ and $x=1$ from where we can conclude that

$$\begin{align} I_n=\frac1n\int_0^{\infty}\frac{e^{-t}t^{-(1-1/n)}}{1+t}dt=\frac1n\frac{\Gamma\left(1-\frac1n,1\right)\Gamma\left(\frac1n\right)}{e^{-1}} \end{align}$$

$$I_n=\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}dx=\frac en\Gamma\left(1-\frac1n,1\right)\Gamma\left(\frac1n\right)$$

Of course this way of solving requires the knowledge of formula $(1)$ $($an impressive proof done by the user Felix Marin can be found within the linked post$)$ but nevertheless this evaluation method is quite compact.

I will offer up (again) a method that coverts the integral to a double integral. Note for convergence of the integral we require $n > 0$.

For $n > 0$, begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives $$I_n = \frac{1}{n} \int_0^\infty \frac{x^{1/n -1} e^{-x}}{1 + x} \, dx \qquad (1)$$

Noting that $$\frac{1}{x + 1} = \int_0^\infty e^{-u(x + 1)} \, du,$$ the integral in (1) can be rewritten as $$I_n = \frac{1}{n} \int_0^\infty x^{1/n - 1} e^{-x} \int_0^\infty e^{-u(x + 1)} \, du \, dx,$$ or $$I_n = \frac{1}{n} \int_0^\infty e^{-u} \int_0^\infty x^{1/n - 1} e^{-x(u + 1)} \, dx \, du,$$ on changing the order of integration.

Enforcing a substitution of $x \mapsto x/(u + 1)$ leads to \begin{align} I_n &= \frac{1}{n} \int_0^\infty (u + 1)^{-1/n} e^{-u} \int_0^\infty x^{1/n - 1} e^{-x} \, dx \, du\\ &= \frac{1}{n} \Gamma \left (\frac{1}{n} \right ) \int_0^\infty (u + 1)^{-1/n} e^{-u} \, du. \end{align} Finally, enforcing a substitution of $u \mapsto u - 1$ one has $$I_n = \frac{e}{n} \Gamma \left (\frac{1}{n} \right ) \int_1^\infty u^{(1 - 1/n) - 1} e^{-u} \, du = \frac{e}{n} \Gamma \left (\frac{1}{n} \right ) \Gamma \left (1 - \frac{1}{n}, 1 \right ),$$ as expected.