Let $\mathrm{I}=\int \sqrt{\frac{x-7}{x-9}} \mathrm{~d} x$
$\begin{aligned}
& =\int \sqrt{\frac{(x-7)(x-7)}{(x-9)(x-7)}} \mathrm{d} x \\
& =\int \frac{x-7}{\sqrt{x^2-16 x+63}} \mathrm{~d} x
\end{aligned}$
Let $(x-7)=\mathrm{A}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(x^2-16 x+63\right)\right]+\mathrm{B}$
$\begin{aligned}
& \therefore \quad x-7=\mathrm{A}(2 x-16)+\mathrm{B} \\
& \therefore \quad x-7=2 \mathrm{~A} x-16 \mathrm{~A}+\mathrm{B} \\
& \therefore \quad \mathrm{A}=\frac{1}{2}, \mathrm{~B}=1 \\
& \therefore \quad \mathrm{I}=\int \frac{\frac{1}{2}(2 x-16)+1}{\sqrt{x^2-16 x+63}} \mathrm{~d} x \\
& =\frac{1}{2} \int \frac{2 x-16}{\sqrt{x^2-16 x+63}} \mathrm{~d} x+\int \frac{1}{\sqrt{x^2-16 x+63}} \mathrm{~d} x \\
& =\frac{1}{2} \times 2 \sqrt{x^2-16 x+63}+\int \frac{1}{\sqrt{(x-8)^2-(1)^2}} \mathrm{~d} x \\
& \ldots\left[\int \frac{\mathrm{f}^{\prime}(x)}{\sqrt{\mathrm{f}(x)}} \mathrm{d} x=2 \sqrt{\mathrm{f}(x)}+\mathrm{c}\right] \\
& \therefore \quad \mathrm{I}=\sqrt{x^2-16 x+63}+\log \left|x-8+\sqrt{x^2-16 x+63}\right|+\mathrm{c} \\
&\end{aligned}$
But, $\int \sqrt{\frac{x-7}{x-9}} \mathrm{~d} x=A \sqrt{x^2-16 x+63}$
$+\log \left|(x-8)+\sqrt{x^2-16 x+63}\right|+\mathrm{c}$
Comparing, we get
$A=1$