Question

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

• A. $8ln\frac{4}{3}$
• B. $8ln\frac{3}{4}$
• C. $4ln\frac{4}{3}$
• D. $0$

Answer 1

Answer: (A)

$8ln\frac{4}{3}$

Given,

$\begin{array}{c}I={\int }_{-1}^1\sqrt{{\left(\frac{x+2}{x-2}\right)}^2+{\left(\frac{x-2}{x+2}\right)}^2}-2dx\\ ={\int }_{-1}^1\sqrt{{\left[\left(\frac{x+2}{x-2}\right)-\left(\frac{x-2}{x+2}\right)\right]}^2}.dx\\ ={\int }_{-1}^1\left|\left(\frac{x+2}{x-2}\right)-\left(\frac{x-2}{x+2}\right)\right|.dx\\ ={\int }_{-1}^1\left|\frac{8x}{x^2-4}\right|.dx[\therefore \frac{8x}{x^2-4}isevenfunction\\ =2{\int }_0^1\left|\frac{8x}{x^2-4}\right|.dx\\ =2{\int }_0^1\frac{-8x}{x^2-4}.dx\\ =-8{\int }_0^1\frac{2x}{x^2-4}.dx\\ =-8\left(\log \left|x^2-4\right|\right)!\\ =-8(log3-log4)\\ \therefore 8\log \left(\frac{4}{3}\right)\\ SothecorrectoptionisA\end{array}$