Question

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

The Correct Option is C

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