If int frac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = sqr | vedclass

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(D) Let $I = \int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}}$.We can rewrite the integrand as $I = \int \frac{dx}{(x-1)(x-1)^{1/2}(x-3)^{1/2}} = \int \frac{dx}

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-$1$

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(D) Let $I = \int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}}$.We can rewrite the integrand as $I = \int \frac{dx}{(x-1)(x-1)^{1/2}(x-3)^{1/2}} = \int \frac{dx}{(x-1)\sqrt{(x-1)(x-3)}}$.Let $x-1 = t$,then $dx = dt$. Also,$x-3 = t-2$.So,$I = \int \frac{dt}{t\sqrt{t(t-2)}} = \int \frac{dt}{t\sqrt{t^2-2t}}$.Substitute $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$.$I = \int \frac{-du/u^2}{(1/u)\sqrt{1/u^2 - 2/u}} = \int \frac{-du/u}{\frac{1}{u}\sqrt{1-2u}} = -\int \frac{du}{\sqrt{1-2u}}$.$I = -\frac{(1-2u)^{1/2}}{1/2 	imes (-2)} + c = (1-2u)^{1/2} + c = \sqrt{1 - \frac{2}{t}} + c = \sqrt{\frac{t-2}{t}} + c$.Substituting $t = x-1$,we get $I = \sqrt{\frac{x-1-2}{x-1}} + c = \sqrt{\frac{x-3}{x-1}} + c$.Thus,$f(x) = \frac{x-3}{x-1}$.Now,$f(-1) = \frac{-1-3}{-1-1} = \frac{-4}{-2} = 2$.And $f(0) = \frac{0-3}{0-1} = \frac{-3}{-1} = 3$.Therefore,$f(-1) - f(0) = 2 - 3 = -1$.

If $\int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = \sqrt{f(x)} + c$,then $f(-1) - f(0) =$

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