$\int\limits_{-1/2}^{1/2}\sqrt{(\frac{x+1}{x-1})^2+(\frac{x-1}{x+1})^2-2}$ dx, equal to:
Answer & explanation
Correct answer: option 4
$\int\limits_{-1/2}^{1/2}\sqrt{(\frac{x+1}{x-1})^2+(\frac{x-1}{x+1})^2-2}$ dx=$\int\limits_{-1/2}^{1/2}\sqrt{(\frac{x+1}{x-1}-\frac{x-1}{x+1})^2}$ dx=$\int\limits_{-1/2}^{1/2}\sqrt{|\frac{x+1}{x-1}-\frac{x-1}{x+1})|^2}$ dx
$I=\int\limits_{-1/2}^{1/2}|\frac{4x}{x^2-1}|dx=2\int\limits_{0}^{1/2}|\frac{4x}{x^2-1}|dx=2\int\limits_{0}^{1/2}\frac{4x}{1-x^2}dx=4\,ln\,(\frac{4}{3})$