If int sqrt{frac{x-7}{x-9}} ~dx = A sqrt{x^2- | vedclass

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(C) Let $I = \int \sqrt{\frac{x-7}{x-9}} ~dx$ Multiply numerator and denominator by $\sqrt{x-7}$: $I = \int \frac{x-7}{\sqrt{(x-9)(x-7)}} ~dx = \int \

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(C) Let $I = \int \sqrt{\frac{x-7}{x-9}} ~dx$ Multiply numerator and denominator by $\sqrt{x-7}$: $I = \int \frac{x-7}{\sqrt{(x-9)(x-7)}} ~dx = \int \frac{x-7}{\sqrt{x^2-16x+63}} ~dx$ Let $x-7 = A \frac{d}{dx}(x^2-16x+63) + B$ $x-7 = A(2x-16) + B$ Comparing coefficients of $x$ and constant terms: $2A = 1 \implies A = \frac{1}{2}$ $-16A + B = -7 \implies -16(\frac{1}{2}) + B = -7 \implies -8 + B = -7 \implies B = 1$ Now,$I = \int \frac{\frac{1}{2}(2x-16) + 1}{\sqrt{x^2-16x+63}} ~dx$ $I = \frac{1}{2} \int \frac{2x-16}{\sqrt{x^2-16x+63}} ~dx + \int \frac{1}{\sqrt{(x-8)^2 - 1^2}} ~dx$ Using $\int \frac{f'(x)}{\sqrt{f(x)}} ~dx = 2\sqrt{f(x)} + c$ and $\int \frac{1}{\sqrt{x^2-a^2}} ~dx = \log |x + \sqrt{x^2-a^2}| + c$: $I = \frac{1}{2} \cdot 2\sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{(x-8)^2 - 1^2}| + c$ $I = 1 \cdot \sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{x^2-16x+63}| + c$ Comparing this with the given expression $A \sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{x^2-16x+63}| + c$,we get $A = 1$.

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

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