Question

sqrt%7B2%7D%7D%7D%5E%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%7D%5Cleft%28%5Cleft%28%5Cfrac%7Bx-1%7D%7Bx%2B1%7D%5Cright%29%5E%7B2%7D%2B%5Cleft%28%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%5Cright%29%5E%7B2%7D-2%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20d%20x%5Cend%7Bequation%7D" id="TexFormula1" title="\begin{equation}\text { Question: If } \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left(\left(\frac{x-1}{x+1}\right)^{2}+\left(\frac{x+1}{x-1}\right)^{2}-2\right)^{\frac{1}{2}} d x\end{equation}" alt="\begin{equation}\text { Question: If } \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left(\left(\frac{x-1}{x+1}\right)^{2}+\left(\frac{x+1}{x-1}\right)^{2}-2\right)^{\frac{1}{2}} d x\end{equation}" align="absmiddle" class="latex-formula">
Options:
(a) ln16
(b) 2ln16
(c) 3ln16
(d) 4ln16

Answer 1

We want to evaluate

$\displaystyle \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

First we note that the integrand is even (replacing x with -x doesn't fundamentally alter the function being integrated), so this is equal to

$\displaystyle 2 \int_0^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

The radicand reduces significantly to

$\displaystyle \left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2 = \frac{16x^2}{(1-x^2)^2}$

so that taking the square root, we simplify the integral to

$\displaystyle 8 \int_0^{\frac1{\sqrt2}} \frac x{1-x^2} \, dx$

which is trivially computed with a substitution of $y = 1 - x^2$ and $dy=-2x\,dx$:

$\displaystyle -4 \int_1^{\frac12} \frac{dy}y = 4 \int_{\frac12}^1 \frac{dy}y = -4 \ln\left(\frac12\right) = \ln(2^4) = \boxed{\ln(16)}$