Question

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

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