Evaluate the integral int frac{1}{sqrt{(x-1)( | vedclass

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(C) We need to evaluate $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$. First,expand the expression inside the square root: $(x-1)(x-2) = x^2 - 3x + 2$. Co

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We need to evaluate $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$. First,expand the expression inside the square root: $(x-1)(x-2) = x^2 - 3x + 2$. Complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4}$. Now the integral becomes $I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx$. Using the standard formula $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}| + C$,where $t = x - \frac{3}{2}$ and $a = \frac{1}{2}$: $I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + C$. Simplifying the expression inside the square root back to the original form: $I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + C$. Thus,the correct option is $C$.

Evaluate the integral $\int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.

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