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

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Aicharya · Accepted Answer

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

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…

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