<s>[INST] The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $ 
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $       

It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it looks like this:

The problem is that I'm not sure that an ellipse here can be a finite area
I will be glad to any hint or solution
 [/INST] Given the function for the ellipse as: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}$$ It is clear that $b$ must be equal to $\frac1c$ so that the ellipse must be below the curve $\frac{\pm1}{x^2+c}$ Solving for $y$ in $(1)$ gives us: $$\left\{y\to\frac{b \sqrt{a^2-x^2}}{a}\right\},\left\{y\to -\frac{b \sqrt{a^2-x^2}}{a}\right\}$$ We know that $b=\frac1c$, therefore, half of the ellipse can be written as: $$y=\frac{\sqrt{a^2-x^2}}{a c}$$ Now, in order for the ellipse to be under the curve, we must ensure that: $$\frac{\sqrt{a^2-x^2}}{a c}<\frac{1}{c+x^2}$$ Which can be rewritten as: $$\frac{1}{c+x^2}-\frac{\sqrt{a^2-x^2}}{a c}>0$$ Now the equality is achieved when for the following solutions in terms of $x$: $$\left\{\{x\to 0\},\left\{x\to -\frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to -\frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\}\right\}$$ Meaning, these are values of $x$ in the $xy$-plane where the ellipse and the curves "intersect". Now, we want this solutions equal to $0$, so that the graphs of the functions only ever intersect at $x=0$ (if that is meaningful). Now, we get the solutions as: $$a\to\pm\frac{\sqrt{c}}{\sqrt{2}}$$ Since $a$ must be greater than 0, we reject the negative answer. Now we have $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, now half of the ellipse is defined by: $$y=\frac{\sqrt{2} \sqrt{\frac{c}{2}-x^2}}{c^{3/2}}$$ Square both sides, and you get the function for the ellipse as: $$y^2=\frac{2 \left(\frac{c}{2}-x^2\right)}{c^3}$$ 

The area the ellipse is given by:
$$A=2\int_{-a}^{a}\frac{b \sqrt{a^2-x^2}}{a}dx$$
Given $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, the maximum area of the ellipse should be:
$$A=2\int_{-\sqrt{\frac{c}{2}}}^{\sqrt{\frac{c}{2}}}\frac{\sqrt2\sqrt{c/2-x^2}}{c^{3/2}}dx=2\frac\pi{2\sqrt{2c}}=\frac\pi{\sqrt{2c}}$$
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