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A friend and I have been swapping difficult integrals for the holidays to stump each other and he recently sent me this one that I haven't been able to figure out (mission accomplished, I guess :) ).

I've tried a few substitutions of the form

$$x-1 = f(t)$$

but if they cancel out one side, they won't simplify on the other because of the presence of both the exponential and the log. At best I could simplify the problem to

$$2 + \int_0^1 e^{1-\frac{1}{x^2}} + \frac{1}{\sqrt{1-\log x}}\:dx$$

by shifting the integral over to the interval $[0,1]$ to see if I could spot any patterns. The integral on the right evaluates to $1$, which is a surprisingly clean answer.

Wolfram gives a complicated looking antiderivative but one of the rules of our little game was that we would invoke no special functions beyond the standard transcendentals and hyperbolics/trig. Even if this was the intended solution, I'm not sure how to simplify the bound at $1$ with the $\operatorname{erf}$s

I suspect he had some clean trick in mind since that was the theme of the game, but I'm stumped.

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Hint: Observe you have \begin{align} I=\int^2_1 e^{1-\frac{1}{(x-1)^2}}+1 +1+\frac{1}{\sqrt{1-\log(x-1)}}\ dx = \int^2_1 [f(x)+f^{-1}(x)]\ dx = 3. \end{align} Draw a picture (for any $f$)!

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    $\begingroup$ Wow that was beautiful! And I feel silly for not spotting that since my friend told me about that trick recently. Cheers! $\endgroup$ – Ninad Munshi Dec 30 '19 at 7:14
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    $\begingroup$ As an edit though, this trick only works if the functions have fixed points at the end points of the integrals. $\endgroup$ – Ninad Munshi Dec 30 '19 at 7:17

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