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:
Correct Option: 2,
Solution:
$I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right)^{1 / 2} d x$
$I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left|\frac{4 x}{x^{2}-1}\right| d x \Rightarrow I=2.4 \int_{0}^{1 / \sqrt{2}}\left|\frac{x}{x^{2}-1}\right| d x$
$\Rightarrow I=-4 \int_{0}^{1 / \sqrt{2}} \frac{2 x}{x^{2}-1} d x \Rightarrow I=-4 \ln \left|x^{2}-1\right|_{0}^{1 / \sqrt{2}}$
$\Rightarrow \mathrm{I}=4 \ln 2 \Rightarrow \mathrm{I}=\ln 16$