Calculate the integral $\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx$

Calculate the integral $\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx$

What I have tried:

Split the integral

$$\int_0^{\infty}\frac{x^{\alpha-1}e^{-x}}{x+2}dx+\int_0^{\infty}\frac{1}{x+2}dx$$

$$=\int_0^{\infty}\frac{x^{\alpha-1}e^{-x}}{x+2}dx$$

Taking the substitution $$u = x+2; du =dx; x= u-2$$ and with the change of variables

$$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du$$

Then integrating by parts with the following substitutions $$z = (u-2)^{\alpha-1}; dx = (\alpha-2)(u-2)^{\alpha-2}du; dv = u^{-1}e^{-u+2}; v = e^2Ei(-u)$$

$$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du=\left[(u-2)^{\alpha-1}e^2Ei(-u)\right]_2^{\infty}-(\alpha-1)e^2\int_2^{\infty}(u-2)^{\alpha-2}Ei(-u)du$$ $$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du=-(\alpha-1)e^2\int_2^{\infty}(u-2)^{\alpha-2}Ei(-u)du$$

How do I proceed in simplifying this any further?

$$\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx>\int_0^\infty\frac1{x+2}=\ln\infty-\ln 2=\infty$$