Question

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

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

Answer 1

The correct option is B $8log\frac{3}{4}$

$I={\int }_{-1}^1{\{{\left(\frac{x+2}{x-2}\right)}^2+{\left(\frac{x-2}{x+2}\right)}^2-2\}}^{\frac{1}{2}}dx={\int }_{-1}^1{\left\{{(\frac{x+2}{x-2}-\frac{x-2}{x+2})}^2\right\}}^{\frac{1}{2}}dx$
$={\int }_{-1}^1\left|\frac{8x}{x^2-4}\right|dx=16{\int }_0^1\left|\frac{x}{x^2-4}\right|dx$
$=16{\int }_0^1\frac{x}{4-x^2}dx=8\{log|4-x^2|{\}}_0^1=8\{log3-log4\}=8log\frac{3}{4}$