Answer
$$\int x^{-5}lnx\,dx=-\frac{lnx}{4x^{4}}-\frac{1}{16x^{4}}+C$$
Work Step by Step
$${(-\frac{x^{-4}lnx}{4})}'=x^{-5}lnx-\frac{x^{-5}}{4}$$
$$\int x^{-5}lnx\,dx=-\frac{x^{-4}lnx}{4}+\frac{1}{4}\int x^{-5}dx$$
$$=-\frac{lnx}{4x^{4}}-\frac{1}{16x^{4}}+C$$
