Question

The value of $\underset{-1/\sqrt{2}}{\overset{1/\sqrt{2}}{\int }}{({\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2)}^{1/2}dx$ is

• A. $2{\log }_{e}16$
• B. ${\log }_{e}16$
• C. $4{\log }_e(3+2\sqrt{2})$
• D. ${\log }_{e}4$

Answer 1

The correct option is B ${\log }_{e}16$

Let $I=\underset{-1/\sqrt{2}}{\overset{1/\sqrt{2}}{\int }}{({\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2)}^{1/2}dx$
$=\underset{-1/\sqrt{2}}{\overset{1/\sqrt{2}}{\int }}{\left({(\frac{x+1}{x-1}-\frac{x-1}{x+1})}^2\right)}^{1/2}dx$
$=\underset{-1/\sqrt{2}}{\overset{1/\sqrt{2}}{\int }}\left|\frac{4x}{x^2-1}\right|dx=2\underset{-1/\sqrt{2}}{\overset{1/\sqrt{2}}{\int }}\frac{2|x|}{1-x^2}dx$

$=2\left\{\underset{-1/\sqrt{2}}{\overset{0}{\int }}\frac{-2x}{1-x^2}dx+{\int }_0^{1/\sqrt{2}}\frac{2x}{1-x^2}dx\right\}$
$=2\{\begin{array}{cc}\ln \begin{array}{c}|1-x^2|\end{array}{|}_{-1/\sqrt{2}}^0& +\begin{array}{c}(-\ln |1-x^2|)\end{array}{|}_0^{1/\sqrt{2}}\end{array}$

$=2\{(0-\ln \frac{1}{2})-(\ln \frac{1}{2}-0)\}=4\ln 2={\log }_{e}16$