definite integrals - Evaluating $int_{0}^{infty} e^{-(x^{n}+x)}, mathrm{d}x,quad n>0$

I know from Discussing the Integral of $\exp(-x^n)$ that $$\int_{0}^{\infty} e^{-x^{n}}\mathrm{d}x=\Gamma(1+1/n),\quad n>0.$$

But how to evaluate $$\int_{0}^{\infty}e^{-(x^{n}-x)}\,\mathrm{d}x,\quad n>0?$$

The only substitution i found is $$\text{Let}\quad x=\ln u, \quad \text{then}\quad e^{x}=u, \quad \text{and} \quad \mathrm{d}x=\frac{1}{u}\mathrm{d}u.$$ Then $$\int_{0}^{\infty}e^{-(x^{n}-x)}\,\mathrm{d}x=\int_{1}^{\infty}e^{-(\ln u)^{n}}\,\mathrm{d}u$$ But after this, I am stuck.

Thank you!