Question

${\int }_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2}dx$ is equal to ?

• A. $4\log \frac{3}{4}$
• B. $4\log \frac{4}{3}$
• C. $2\log \frac{16}{9}$
• D. $-\log \frac{81}{256}$

Answer 1

The correct option is B $4\log \frac{4}{3}$

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

As $\left|\frac{4x}{x^2-1}\right|$ is an even function therefore from the property of definite integrals.

$I=2{\int }_0^{1/2}\left|\frac{4x}{x^2-1}\right|dx=2\times 2{\int }_0^{1/2}\frac{-2x}{x^2-1}dx=-4{\left[log[x^2-1]\right]}_0^{1/2}=4log\frac{4}{3}$