For which p one can calculate this improper integral $\displaystyle \int_{3}^{\infty}\frac{x^p(1-e^{-x})}{1+x^p}dx $?(the "3" is not important!)
I tried $$\int_{3}^{\infty}\frac{x^p(1-e^{-x})}{1+x^p}dx=\int_{3}^{\infty}\frac{(x^p+1-1)(1-e^{-x})}{1+x^p}dx=\\=\int_{3}^{\infty}(1-e^{-x})dx-\int_{3}^{\infty}\frac{(1-e^{-x})}{1+x^p}dx=... $$
But I didn't succeed to continue.