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It's obvious that there exists $p(x)>1$ such that $$\frac1{(1+|x|)^{p(x)}}=\frac12\frac1{1+|x|}\quad\quad(x\ne0).$$

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OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula. $$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first ...

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One such condition would be: there is a countable set $C$ such that$$(\forall x\in\mathbb R\setminus C):f(x)\geqslant g(x).$$

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Before attacking the integral, I mention something about cubic theta function. The whole solution is quite self-contained if you accept the stated facts, the "footnote" contains more information. The three cubic theta functions are defined by \begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) ... 1 Let u(x) = f'(x) and v(x) = \sqrt{1+u(x)^2}. So we want u to have an elementary integral (so that we can write down f on the assignment sheet) and v to have an elementary integral (so our students can solve it.) In other words, we want functions u and v, both with elementary integrals, so that v^2 = 1 + u^2. Rewrite this as (v+u) (v-u) = 1.... 0 Your error lies in the computation of the intersection points of the curves. If y^2=4x and y=4x-2, then 4x=(4x-2)^2 indeed, but this is not an equivalence. It turns out that your region is the region below the graph of 2\sqrt x and above the graph of -2\sqrt x (with x\in\left[0,\frac14\right]) plus the area of the region below the graph of 2\... 1 If you actually look at a graph of the required area you will see that it is quite difficult to find when integrating with respect to x. We can instead rearrange both equations to getx=\frac14y^2x=\frac14y+\frac12$$Then the graphs intersect at points where y=-1 and y=2 respectively so the area is given by$$\int_{-1}^2\left(\frac14y+\frac12\...

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Here is an independent solution: \begin{align} I&=\int_0^1\frac{\ln y\ln^2(1+y)}{y}\ dy\overset{y=\frac{1-x}{x}}{=}\int_{1/2}^1\frac{\ln(1-x)\ln^2x-\ln^3x}{x(1-x)}\ dx\\ &=\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln^2x}{x}\ dx}_{IBP}-\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^3x}{x}\ dx}_{-(\ln^42)/4}+\int_{1/2}^1\frac{\ln(1-x)...

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