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

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Question

(C) Let $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}} = \int \frac{dx}{\sqrt{x^2-3x+2}}$.By completing the square in the denominator:$x^2-3x+2 = (x^2-3x+\fra

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$\cosh ^{-1}(2x-3)+c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}} = \int \frac{dx}{\sqrt{x^2-3x+2}}$.By completing the square in the denominator:$x^2-3x+2 = (x^2-3x+\frac{9}{4}) - \frac{9}{4} + 2 = (x-\frac{3}{2})^2 - \frac{1}{4} = (x-\frac{3}{2})^2 - (\frac{1}{2})^2$.Thus, $I = \int \frac{dx}{\sqrt{(x-\frac{3}{2})^2 - (\frac{1}{2})^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}(\frac{x}{a}) + c$:$I = \cosh^{-1}(\frac{x-3/2}{1/2}) + c = \cosh^{-1}(2x-3) + c$.

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

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