If int sqrt{frac{x - 5}{x - 7}} dx = A sqrt{x | vedclass

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(D) To evaluate $\int \sqrt{\frac{x - 5}{x - 7}} dx$,we multiply the numerator and denominator by $\sqrt{x - 5}$:$\int \frac{x - 5}{\sqrt{(x - 5)(x -

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$1$

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(D) To evaluate $\int \sqrt{\frac{x - 5}{x - 7}} dx$,we multiply the numerator and denominator by $\sqrt{x - 5}$:$\int \frac{x - 5}{\sqrt{(x - 5)(x - 7)}} dx = \int \frac{x - 5}{\sqrt{x^2 - 12x + 35}} dx$Next,we express the numerator in terms of the derivative of the quadratic expression $x^2 - 12x + 35$,which is $2x - 12$:$= \frac{1}{2} \int \frac{2x - 10}{\sqrt{x^2 - 12x + 35}} dx = \frac{1}{2} \int \frac{2x - 12 + 2}{\sqrt{x^2 - 12x + 35}} dx$$= \frac{1}{2} \int \frac{2x - 12}{\sqrt{x^2 - 12x + 35}} dx + \int \frac{1}{\sqrt{x^2 - 12x + 35}} dx$$= \sqrt{x^2 - 12x + 35} + \int \frac{1}{\sqrt{(x - 6)^2 - 1}} dx$Using the standard integral $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}|$,we get:$= \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{(x - 6)^2 - 1}| + C$$= 1 \cdot \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{x^2 - 12x + 35}| + C$Comparing this with the given form,we find $A = 1$.

If $\int \sqrt{\frac{x - 5}{x - 7}} dx = A \sqrt{x^2 - 12 x + 35} + \log |x - 6 + \sqrt{x^2 - 12 x + 35}| + C$,then $A = . . . . . .$

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