Question

`\begin{equation}\text { Question: If } \int_{(-1)/(√(2))}^{(1)/(√(2))}\left(\left((x-1)/(x+1)\right)^(2)+\left((x+1)/(x-1)\right)^(2)-2\right)^{(1)/(2)} d x\end{equation}`

Options:
(a) ln16
(b) 2ln16
(c) 3ln16
(d) 4ln16

Answer 1

We want to evaluate

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

The radicand reduces significantly to

`\displaystyle \left((x-1)/(x+1)\right)^2 + \left((x+1)/(x-1)\right)^2 - 2 = (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)}`