If
\(\int \sqrt{\frac{x-7}{x-9}} \mathrm{~d} x=A \sqrt{x^2-16 x+63}\) \(+\log \left|(x-8)+\sqrt{x^2-16 x+63}\right|+c,\) (where \(\mathrm{c}\) is a constant of integration) then \(\mathrm{A}\) is

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 \((x-7)=\mathrm{A}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(x^2-16 x+63\right)\right]+\mathrm{B}\)
\(\therefore x-7=\mathrm{A}(2 x-16)+\mathrm{B} \)
\( \therefore x-7=2 \mathrm{~A} x-16 \mathrm{~A}+\mathrm{B} \)
\( \therefore \mathrm{A}=\frac{1}{2}, \mathrm{~B}=1 \)
\( \therefore \mathrm{I}=\int \frac{\frac{1}{2}(2 x-16)+1}{\sqrt{x^2-16 x+63}} \mathrm{~d} x \)
\( =\frac{1}{2} \int \frac{2 x-16}{\sqrt{x^2-16 x+63}} \mathrm{~d} x+\int \frac{1}{\sqrt{x^2-16 x+63}} \mathrm{~d} x \)
\( =\frac{1}{2} \times 2 \sqrt{x^2-16 x+63}+\int \frac{1}{\sqrt{(x-8)^2-(1)^2}} \mathrm{~d} x \)
\( \ldots\left[\int \frac{\mathrm{f}^{\prime}(x)}{\sqrt{\mathrm{f}(x)}} \mathrm{d} x=2 \sqrt{\mathrm{f}(x)}+\mathrm{c}\right] \)
\( \therefore \mathrm{I}=\sqrt{x^2-16 x+63}+\log |x-8+\) \(\sqrt{x^2-16 x+63}|+\mathrm{c}\)
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
\(A=1\)