约分：x^2y-2xy^2分之3x^3-6x^2y其中x=3.y=1, 约分：x^2y-2xy^2分之3x^3-6x^2
约分：x^2y-2xy^2分之3x^3-6x^2y其中x=3.y=1 问题补充：
5a^2b+ab^2分之25a^2-b^2,其中a=-0.2,b=10 半夏Tiramisu 约分：x^2y-2xy^2分之3x^3-6x^2y其中x=3.y=1
(1) (3x^3-6x²y)/(x²y-2xy²)=3x²(x-2y)/[xy(x-2y)]
当 x=3，y=1
原式=9(2) (25a²-b²)/(5a²b+ab²)=(5a-b)(5a+b)/[ab(5a+b)]
=(5a-b)/(ab)
a=-0.2，b=10
原式=(-2-10)/(-2)
（1）x^2y-2xy^2分之3x^3-6x^2y=xy(x-2y）分之3x²(x-2y)=y分之3x当x=3.y=1时，x^2y-2xy^2分之3x^3-6x^2y==y分之3x=9（2）5a^2b+ab^2分之25a^2-b^2=ab（5a+b)分之(5a+b)(5a-b)=ab分之(5a-b)=2分之11calculus - Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$ - Mathematics Stack Exchange
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I need help with this integral:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
The integrand graph looks like this:
$\hspace{1in}$
The approximate numeric value of the integral:
$$I\approx8.121...$$
Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in
did not return plausible closed form candidates either. But I still hope there might be a closed form for it.
I am also interested in cases when only numerator or only denominator is present under the logarithm.
4,58941748
3,02433954
I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists.
From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand.
First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get
$$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$
Now use the symmetry from the map $t \mapsto 1/t$.
Break the integral up into two as follows:
\begin{align}
& 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )}
\end{align}
Sub $t=u^2$ to get
$$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$
Integrate by parts:
$$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$
One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get
$$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$
With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it.
To wit, consider the following contour integral:
$$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$
where $C$ is a keyhole contour about the positive real axis.
This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs)
$$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$
It should be noted that the seco this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$.
On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand.
In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically.
Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root.
For example, we may deduce that
$$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1)
(z^4-4 z^3+10 z^2-4 z+1)$$
which exploits the $a \mapsto -a$ symmetry.
$$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$
Write $a=r e^{i \theta}$ and get the following equations:
$$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$
$$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$
From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio.
Thus the poles take the form
$$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$
Now we have to find the residues of the integrand at these 8 poles.
We can break this task up by computing:
$$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)
\log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$
Here things got very messy, but the result is rather unbelievably simple:
$$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$
Actually, this is a very simple computation.
Inspired by @sos440, one may express the rational function of $z$ in a very simple form:
$$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$
$$p(z)=z^4+4 z^3+10 z^2+4 z+1$$
The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$.
That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$.
Now consider the log piece.
Expanding the square, we get 3 terms:
$$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$
Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero.
This leaves the second term.
For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$.
Thus, we have
$$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$
12.9k71854
107k12129218
NEW ANSWER. I found yet another way of solving this problem. My new solution does not use contour integration, and is based on the following observation: for $|z| \leq 1$,
$$ - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . $$
As I want to keep both the old answer and the new answer, I posted my new solution to other page. You can check it .
OLD ANSWER. Okay here is another solution. It is also related to .
We claim the following proposition:
Proposition. If $0 & r & 1$ and $r & s$, then
$$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{1} $$
Assuming this proposition, all that we have to do is to solve the non-linear system of equations
$$ 2rs = 2 \quad \text{and} \quad r^{2} + s^{2} - 1 = 2. $$
The unique solution satisfying the condition of the proposition is $r = \phi - 1$ and $s = \phi$. So by $\text{(1)}$ we have
\begin{align*}
\int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2x + 2x^{2}}{1 - 2x + 2x^{2}} \right) \, dx
& = I(\phi-1, \phi) \\
&= 4\pi \arcsin (\phi - 1)
= 4\pi \operatorname{arccot} \sqrt{\phi}.
\end{align*}
Thus it remains to prove the proposition.
Proof of Proposition. We divide the proof into several steps.
Step 1. (Case reduction by analytic continuation) We first remark that given $r$ and $s$, we always have
$$ \min_{-1 \leq x \leq 1} \{ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} \} & 0. \tag{2} $$
Indeed, it is not hard to check if we utilize the following equality
$$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm rsx)^{2} - (1 - r^{2})(1 - s^{2}) x^{2}. $$
Then $\text{(2)}$ shows that the integrand of $I(r, s)$ remains holomoprhic under small perturbation of $s$ in $\Bbb{C}$. So it allows us to extend $s \mapsto I(r, s)$ as a holomorphic function on some open set containing the line segment $(r, \infty) \subset \Bbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r & s & 1$.
Step 2. (Integral representation of $I$) Put $r = \sin \alpha$ and $s = \sin \beta$, where $ 0 & \alpha & \beta & \frac{\pi}{2}$. Then
\begin{align*}
&= \int_{-1}^{1} \frac{1+x}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \\
&= \int_{0}^{1} \frac{2}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \qquad (\because \text{ parity}) \\
&= \int_{1}^{\infty} \frac{2}{\sqrt{x^{2}-1}} \log \left( \frac{x^{2} + 2rsx + (r^{2} + s^{2} - 1)}{x^{2} - 2rsx + (r^{2} + s^{2} - 1)} \right) \, dx \qquad (x \mapsto x^{-1}) \\
&= \int_{0}^{1} \frac{2}{t} \log \left( \frac{\left(t+t^{-1}\right)^{2} + 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)}{\left(t+t^{-1}\right)^{2} - 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)} \right) \, dt,
\end{align*}
where in the last line we utilized the substitution $x = \frac{1}{2}(t + t^{-1})$. If we introduce the quartic polynomial
\begin{align*}
p(t) = t^{4} + 4rst^{3} + (4r^{2}+4s^{2}-2)t^{2} + 4rst + 1,
\end{align*}
then by the property $p(1/t) = t^{-4}p(t)$, we can simplify
\begin{align*}
&= 2 \int_{0}^{1} \frac{\log p(t) - \log p(-t)}{t} \, dt
= \int_{0}^{\infty} \frac{\log p(t) - \log p(-t)}{t} \, dt \\
&= - \int_{0}^{\infty} \left( \frac{p'(t)}{p(t)} + \frac{p'(-t)}{p(-t)} \right) \log t \, dt
= - \frac{1}{2} \Re \int_{-\infty}^{\infty} \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z \, dz,
\end{align*}
where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane
$$\Bbb{H} = \{ z \in \Bbb{C} : \Im z & 0 \}.$$
Step 3. (Residue calculation) Since
$$ f(z) := \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z = O\left(\frac{\log z}{z^{2}} \right) \quad \text{as } z \to \infty, $$
by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that
\begin{align*}
I(r, s) = - \frac{1}{2} \Re \left\{ 2 \pi i \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z) \right\} = \pi \Im \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z).
\end{align*}
(It turns out that $f(z)$ has only logarithmic singularity at the origin. So it does not account for the value of $I(r, s)$.) But by a simple calculation, together with the condition $ 0 & \alpha & \beta & \frac{\pi}{2}$, we easily notice that the zeros of $p(z)$ are exactly
$$ e^{\pm i(\alpha + \beta)} \quad \text{and} \quad -e^{\pm i(\alpha - \beta)}. $$
Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\Bbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\Bbb{H}$. Then
$$ Z_{+} = \{ e^{i(\beta+\alpha)}, -e^{-i(\beta - \alpha)} \} \quad \text{and} \quad Z_{-} = \{ e^{-i(\beta+\alpha)}, -e^{i(\beta- \alpha)} \}. $$
This in particular shows that
$$ \frac{p'(z)}{p(z)}\log z = \sum_{z_{0} \in Z_{+}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H} $$
$$ \frac{p'(-z)}{p(-z)}\log z = -\sum_{z_{0} \in -Z_{-}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H}. $$
So it follows that
\begin{align*}
&= \pi \Im \left\{ \sum_{z_{0} \in Z_{+}} \log z_{0} - \sum_{z_{0} \in -Z_{-}} \log z_{0} \right\} \\
&= \pi \Im \left\{ \log e^{i(\beta+\alpha)} + \log e^{i(\pi-\beta+\alpha)} - \log e^{i(\pi-\beta-\alpha)} - \log e^{i(\beta-\alpha)} \right\} \\
&= \pi \Im \left\{ i(\beta+\alpha) + i(\pi-\beta+\alpha) - i(\pi-\beta-\alpha) - i(\beta-\alpha) \right\} \\
&= 4\pi \alpha
= 4\pi \arcsin r.
\end{align*}
This completes the proof.
51.2k6103177
$\large\hspace{3in}I=4\,\pi\operatorname{arccot}$
Our aim is to give an elementary proof of Proposition formula (1) in the answer of @sos440.
We first note that
\min_{-1\leq x\leq1}\{1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\}&0.
Indeed, if $x=\pm1$ then
1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\geq(r-s)^{2}&0,
if $x=0$ then
1\pm2rsx+(r^{2}+s^{2}-1)x^{2}=1&0,
if $-1&x&1$, $x\neq0$ then the equations
\begin{eqnarray*}
\frac{\partial}{\partial s}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0,\\
\frac{\partial}{\partial r}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0,
\end{eqnarray*}
give $\pm r=sx$, $\pm s=rx$, which is impossible.
In the second step we show that $I(r,s)$ is independent of $s$.
\frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx.
Substituting $x:=-x$ and adding them we obtain
2\frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{2}{\sqrt{1-x^{2}}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx,
\frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{1}{\sqrt{1-x^{2}}}\cdot\frac{4r(-s^{2}+r^{2}-1)x^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}x^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)x^{2}}\, dx.
Substituting $x:=\sin(t)$ we have
\frac{\partial}{\partial s}I(r,s)
\int_{-\pi/2}^{\pi/2}\frac{4r(-s^{2}+r^{2}-1)\sin(t)^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}\sin(t)^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)\sin(t)^{2}}\, dt
=\int_{-\pi/2}^{\pi/2}-\frac{8r((-s^{2}+r^{2}-1)\cos(2t)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(2t)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(2t)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dt
= \int_{-\pi}^{\pi}-\frac{4r((-s^{2}+r^{2}-1)\cos(y)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(y)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(y)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dy.
Introducing the new variable $T:=\tan\frac{y}{2}$ we obtain
\begin{eqnarray*}
\frac{\partial}{\partial s}I(r,s) & = & \int_{-\infty}^{\infty}-\frac{4r(s^{2}-r^{2})T^{2}-4r}{(r-s)^{2}(r+s)^{2}T^{4}+((2-4s^{2})r^{2}+2s^{2})T^{2}+1}\, dT\\
& = & -\frac{4r(s^{2}-r^{2})}{(r-s)^{2}(r+s)^{2}}\int_{-\infty}^{\infty}\frac{T^{2}+a}{T^{4}+bT^{2}+b^{2}/4+d}\, dT\\
& = & -\frac{4r(-s^{2}+r^{2})}{(r-s)^{2}(r+s)^{2}}\cdot\frac{(2a(b^{2}+4d)+(b^{2}+4d)^{3/2})\pi}{(b^{2}+4d)^{3/2}\sqrt{\sqrt{b^{2}+4d}+b}},
\end{eqnarray*}
a=-\frac{1}{s^{2}-r^{2}},
b=\frac{(2-4s^{2})r^{2}+2s^{2}}{(r-s)^{2}(r+s)^{2}},
b^{2}+4d=\frac{4}{(r-s)^{2}(r+s)^{2}}.
It gives $2ab^{2}+8da+(b^{2}+4d)^{3/2}=0$.
Since $\frac{\partial}{\partial s}I(r,s)=0$ we have
I(r,s)=I(r,1)=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{(1+rx)^{2}}{(1-rx)^{2}}\right)dx.
\frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\frac{4}{1-r^{2}x^{2}}\, dx.
Similarly as above we get
\frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\frac{4}{\sqrt{1-x^{2}}(1-r^{2}x^{2})}\, dx=\frac{4\pi}{\sqrt{1-r^{2}}}=4\pi(\arcsin r)'.
It implies
I(r,1)=4\pi\arcsin r+C.
Taking the limit $\lim_{r\to0+}$ we obtain $C=0$, that is, $I(r,s)=4\pi\arcsin r$.
For the purposes of alternative methods, it may be of interest to note that the integrand
$$f(x)=\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)$$ may be rewritten in terms of hyperbolic trigonometric functions. Using $$\tan^{-1}(z) = \frac{i}{2}\log\left(\frac{1-iz}{1+iz}\right),$$ we obtain
$$f(x)=\frac{1}{x}e^{\tanh^{-1}x}\log\left(\frac{1+\frac{2x}{1+2x^2}}{1-\frac{2x}{1+2x^2}}\right) = e^{\tanh^{-1} x}\left(\frac{2\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)}{x}\right).$$
The rational function in the bracket, which we will denote $s(x)$, is symmetric about $x=0$.
The desired integral is
$$I=\int_{-1}^1 f(x)dx = \int_{-1}^1e^{\tanh^{-1}x}s(x)dx,$$
which, by adding the indicated useful definite integral to both side, gives
$$I + \int_{-1}^1 e^{-\tanh^{-1}x}s(x)dx = 2\int_{-1}^1 \frac{s(x)dx}{\sqrt{1-x^2}}.$$
Now using the change of variable $x=-y$ we have $$\int_{-1}^1 e^{-\tanh^{-1} x}s(x)dx = -\int_1^{-1} e^{\tanh y}s(-y)dy = \int_{-1}^1 e^{\tanh y}s(y)dy = I,$$ by the symmetry of $s(x)$. Hence, we finally obtain
$$I = \int_{-1}^1\frac{s(x)dx}{\sqrt{1-x^2}} = 2\int_{-1}^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
This integral is symmetric about $x=0$, so we have
$$I=4\int_0^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx,$$ which can be rewritten $$I=-4\int_0^1\left(\frac{d}{dx}\text{sech}^{-1}x\right)\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
Using integration by parts this results in
$$I=8\int_0^1\frac{\text{sech}^{-1}(x)(1-2x^2)}{1+4x^4}dx.$$
We can also make the
change of variable $y=\text{sech}^{-1}x$ to obtain
$$I=8\int_0^\infty\frac{y(\cosh^2(y)-2)\sinh y}{\cosh^4(y)+4}dy= 8\int_0^\infty\frac{y\sinh^3 y}{\cosh^4y+4}dy-8\int_0^\infty\frac{y\sinh y}{\cosh^4 y+4}dy.$$
Noteworthy, RIES () finds closed form from numerical value in the form of an equation:
\cos{\left( \frac{x}{\pi} \right)}+1=\frac{2}{\phi^6}.
Simplifying above, we get another form of the result:
I = \pi \arccos{(17-8\sqrt{5})}.
This is not really an answer, but grossly too long for an comment. I didn't know how to simplify it beyond the final solution.
$$I=\int_{-1}^1 \frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\text{d}{x}$$
Begin with the substitution of $x=-\cos2a$
$$I=\int_{-1}^1 \frac{1}{-\cos2a}\sqrt{\frac{1-\cos2a}{1+\cos2a}}\ln\left(\frac{2\cos^2 2a-2\cos 2a+1}{2\cos^2 2a-2\cos2a+1}\right)\text{d}{x}$$
By the tangent and cos double angle properties
$$I=\int_{-1}^1 -\sec2a|\tan a|\ln\left(\frac{-2\cos^22a+\cos 4a+2}{2\cos2a+\cos4a+2}\right)\text{d}{a}$$
Were just getting started. Now replace $a=\frac{1}{2}\text{gd}(b)$ where $\text{gd}$ is the Gudermannian function.
$$I=\int_{-1}^1 -\sec(\text{gd}(b))|\tan(\text{gd}(\frac{b}{2}))|\ln\left(\frac{-2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}{2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}\right)\text{d}{a}$$
Hehe. Now we get to simplify a bit. This is under the definition of Gudermannian properties.
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{-2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}{2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}\right)$$
Now, use properties of $\tanh$ and $\text{sech} $ to simplify even further
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{(1-\text{sech}^2 b)+2}{(1+\text{sech}^2 b)+2}\right)$$
Our goal is to create an $\text{arctanh}$ function, but that will obviously take some serious effort. Factor out a $3$ to generate that $1$ needed even if it makes an ugly factoring.
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{3(1-\frac{\text{sech}^2 b}{3})}{3(1+\frac{\text{sech}^2 b}{3})}\right)$$
And now cut out all of the 3's. After this cut, use a property of $\ln$'s to reciprocate the argument of $\ln$. And multiply 2 and 1/2
$$I=\int_{-1}^1 2\text{cosh}\space b|\sinh\frac{b}{2}|\frac{1}{2}\ln\left(\frac{(1+\frac{\text{sech}^2 b}{3})}{(1-\frac{\text{sech}^2 b}{3})}\right)$$
And what do you know! You're there! Use a property of $\ln$ and $\text{arctanh}$ to generate a much CLEANER form (also by throwing the 2 in front).
$$I=2\int_{-1}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$
This function is even, and we can know that because all parts of what is above, $\cosh b,|\sinh b|, $ etc. all even. So we can do the following.
$$I=4\int_{0}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$
This is just an idea, and like I said not a real solution. I have no idea where to continue beyond this, but I thought it may help to come up with a new idea to solve.
user285523
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