The value of
Question:

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:

  1. $\log _{e} 4$

  2. $\log _{\mathrm{e}} 16$

  3. $2 \log _{e} 16$

  4. $4 \log _{\mathrm{e}}(3+2 \sqrt{2})$


Correct Option: 2,

Solution:

$I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right)^{1 / 2} d x$

$I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left|\frac{4 x}{x^{2}-1}\right| d x \Rightarrow I=2.4 \int_{0}^{1 / \sqrt{2}}\left|\frac{x}{x^{2}-1}\right| d x$

$\Rightarrow I=-4 \int_{0}^{1 / \sqrt{2}} \frac{2 x}{x^{2}-1} d x \Rightarrow I=-4 \ln \left|x^{2}-1\right|_{0}^{1 / \sqrt{2}}$

$\Rightarrow \mathrm{I}=4 \ln 2 \Rightarrow \mathrm{I}=\ln 16$

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