Ytukyg

Is this conjecture true?

Conjecture:
$$
int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8}
$$

I found it myself based on numerical evidence. Need help in analytical proof. Thanks.

It is easier than I exspected in the first place. However, first of all lets denote your integral as

$$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$

We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as

$$begin{align*}
mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
end{align*}$$

Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to

$$begin{align*}
mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
&=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
&=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
&=frac12left[arctan{z}right]_{-infty}^1
end{align*}$$

$$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$

Where we used the symmetry aswell as well-known values of the tangent function.

Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
$$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.

The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.

You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.