Integral #3

$\begin{aligned} f(t) & = \int_2^t \frac 1{1-x^2} dx \\ & = \int_2^t \frac 1{(x-1)(x+1)} dx \\ & = \frac 12 \int_2^t \left(\frac 1{1-x}+\frac 1{1+x}\right) dx \\ & = \frac 12 \ln \left(\frac {x+1}{x-1} \right) \bigg|_2^t \\ & = \frac 12 \left(\ln \left(\frac {t+1}{t-1} \right) - \ln 3 \right) \\ & = \frac 12 \ln \left(\frac {t+1}{3(t-1)} \right) \end{aligned}$

Then we have:

$\begin{aligned} \exp \left(\lim_{t \to \infty} - f(t)\right) & = \lim_{t \to \infty} \exp \left(- f(t)\right) \\ & = \lim_{t \to \infty} \exp \left(-\frac 12 \ln \frac {t+1}{3(t-1)} \right) \\ & = \lim_{t \to \infty} \sqrt{\frac {3(t-1)}{t+1}} & \small \blue{\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \lim_{t \to \infty} \sqrt{\frac 31} & \small \blue{\text{Differentiate up and down w.r.t. }t.} \\ & = \sqrt 3 \end{aligned}$

Area of an equilateral triangle is given by $\dfrac {\sqrt 3 s^2}4 = \sqrt 3 \implies s = \boxed 2$ .

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Reference: L'Hôpital's rule

The given integral equals $\dfrac{1}{2}(\ln |t+1|-\ln |t-1|-\ln 3)$ . As $t$ tends to infinity, $\ln |\dfrac{t+1}{t-1}|$ approaches $\ln 1=0$ , and $f(t)$ approaches $-\ln \sqrt 3$ and $\exp \left (-f(t)\right)$ approaches $\sqrt 3$ . So $\sqrt 3=\dfrac{\sqrt 3s^2}{4}$ and hence $s=\boxed 2$ .