int frac{dx}{sqrt{(x-1)(x-2)}}= | vedclass

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(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.Now,complete th

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$\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3 x+2}\right|+c$

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(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.Now,complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.The integral becomes $I = \int \frac{dx}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$,we get:$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + c$.Simplifying the expression inside the square root back to the original form,we get:$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + c$.

$\int \frac{dx}{\sqrt{(x-1)(x-2)}}=$

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