The value of the integral int_{-1 / 2}^{1 / 2 | vedclass

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(C) Let $I = \int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x$.Using the identity $a^2 + b^2 - 2 =

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$4 \log _{e}\left(\frac{4}{3}\right)$

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(C) Let $I = \int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x$.Using the identity $a^2 + b^2 - 2 = (a-b)^2$,we have $\sqrt{(a-b)^2} = |a-b|$.Thus,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{x+1}{x-1} - \frac{x-1}{x+1} \right| d x$.Simplifying the expression inside the absolute value: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{4x}{x^2-1}$.So,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.Since the integrand is an even function,$I = 2 \int_{0}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.For $x \in [0, 1/2]$,$x^2-1 $I = 2 \int_{0}^{1 / 2} \frac{4x}{1-x^2} d x = 4 \int_{0}^{1 / 2} \frac{2x}{1-x^2} d x$.Let $u = 1-x^2$,then $du = -2x dx$.$I = 4 [-\ln|1-x^2|]_{0}^{1/2} = 4 [-\ln(3/4) + \ln(1)] = 4 \ln(4/3)$.

The value of the integral $\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x$ is equal to

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