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

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Question

(B) Let $I = \int_{-1 / \sqrt{2}}^{1 / \sqrt{2}} \sqrt{\left(\frac{x+1}{x-1}\right)^{2} + \left(\frac{x-1}{x+1}\right)^{2} - 2} \, dx$. We know that $

Vedclass · Accepted Answer

$\log _{e} 16$

Vedclass · Answer

(B) Let $I = \int_{-1 / \sqrt{2}}^{1 / \sqrt{2}} \sqrt{\left(\frac{x+1}{x-1}\right)^{2} + \left(\frac{x-1}{x+1}\right)^{2} - 2} \, dx$. We know that $a^2 + b^2 - 2 = (a-b)^2$. Here $a = \frac{x+1}{x-1}$ and $b = \frac{x-1}{x+1}$. So,the integrand becomes $\sqrt{(\frac{x+1}{x-1} - \frac{x-1}{x+1})^2} = |\frac{x+1}{x-1} - \frac{x-1}{x+1}|$. Simplifying the expression inside the absolute value: $\frac{(x+1)^2 - (x-1)^2}{x^2-1} = \frac{4x}{x^2-1}$. Thus,$I = \int_{-1 / \sqrt{2}}^{1 / \sqrt{2}} |\frac{4x}{x^2-1}| \, dx$. Since the integrand is an even function,$I = 2 \int_{0}^{1 / \sqrt{2}} |\frac{4x}{x^2-1}| \, dx = 8 \int_{0}^{1 / \sqrt{2}} |\frac{x}{x^2-1}| \, dx$. For $x \in [0, 1/\sqrt{2}]$,$x^2-1 $I = -8 \int_{0}^{1 / \sqrt{2}} \frac{x}{x^2-1} \, dx = -4 \int_{0}^{1 / \sqrt{2}} \frac{2x}{x^2-1} \, dx$. $I = -4 [\ln |x^2-1|]_{0}^{1 / \sqrt{2}} = -4 (\ln |1/2 - 1| - \ln |0-1|) = -4 (\ln(1/2) - 0) = -4 \ln(1/2) = 4 \ln 2 = \ln 16$.

The value of $\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:

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