Question

The value of the integral ${\int }_{-1/2}^{1/2}{[{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2]}^{1/2}dx$ is

• A. $\log \left(\frac{4}{3}\right)$
• B. $4\log \left(\frac{3}{4}\right)$
• C. $4\log \left(\frac{4}{3}\right)$
• D. $\log \left(\frac{3}{4}\right)$

Answer 1

Answer: (C)

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

${\int }_{-1/2}^{1/2}{[{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2]}^{1/2}dx$
$={\int }_{-1/2}^{1/2}{[{(\frac{x+1}{x-1}+\frac{x-1}{x+1})}^2-2]}^{1/2}dx$
$={\int }_{-1/2}^{1/2}\left|\frac{4x}{x^2-1}\right|dx$
$={\int }_{-1/2}^0\left|\frac{4x}{1-x^2}\right|dx+{\int }_0^{1/2}\left|\frac{4x}{1-x^2}\right|dx$
$=-4{\int }_{-1/2}^0\frac{x}{1-x^2}dx+4{\int }_0^{1/2}\frac{x}{1-x^2}dx$
$=2{\left\{\log \right(1-x^2\left)\right\}}_{-1/2}^0-2{\left\{\log \right(1-x^2\left)\right\}}_0^{1/2}$
$=-2\log (1-\frac{1}{4})-2\log (1-\frac{1}{4})$
$=-4\log \frac{3}{4}=4\log \frac{4}{3}$

Hence, ${\int }_{-1/2}^{1/2}{[{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2]}^{1/2}dx=4\log \frac{4}{3}$