$I = \int\limits_1^9 {\frac{{ln(16 - x)}}{{\sqrt x }}} dx$Đặt: \[u = ln(16 - x) \Rightarrow du = \frac{{dx}}{{x - 16}}\]
\[dv = \frac{{dx}}{{\sqrt x }} \Leftarrow v = 2\sqrt x \]
$ \Rightarrow I = 2\sqrt x ln(16 - x) - 2\int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx = 6\ln 7 - 2\ln 15 - 2\int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx$
$K = \int\limits_1^9 {\frac{{\sqrt x }}{{x - 16}}} dx = \int\limits_1^9 {\frac{{x - 16 + 16}}{{\sqrt x (x - 16)}}} dx = \int\limits_1^9 {\frac{1}{{\sqrt x }}} dx + \int\limits_1^9 {\frac{{16}}{{\sqrt x (x - 16)}}} dx$
$ = 2\sqrt x\begin{cases}9 \\ 1 \end{cases} + \int\limits_1^9 {\frac{{16}}{{\sqrt x (x - 16)}}} dx$
Đặt $t = \sqrt x \Rightarrow 2dt = \frac{{dx}}{{\sqrt x }}$
$ \Rightarrow K = 4 + \int\limits_1^3 {\frac{{32}}{{{t^2} - 16}}} dt = 4 + \int\limits_1^3 {(\frac{4}{{t - 4}}} - \frac{4}{{t + 4}})dt $
$= 4 + 4ln\left| {\frac{{t - 4}}{{t + 4}}} \right|\begin{cases}3 \\ 1 \end{cases} = 4 + 4ln\frac{5}{{21}}$
Vậy \[I = 6\ln 7 - 2\ln 15 - 4 - 8ln\frac{5}{{21}}\]