$\huge \begin{align} \int_{5}^{10} \frac{8}{x-1}\d x &= 10\int_{5}^{10} (x-1)^{-1}\d x \\ \\ \let u &= x-1\\ \let \d u &= \d x \\ \\ &= 8\int_{5}^{10} u^{-1}\d u \\ &= 8\pa{\ln (x-1) \big\rvert_{5} ^{10}} \\ &= 8\pa{\ln(9)-\ln 5}\\ &= \boxed{7\ln \frac{9}{5}} \end{align}$