Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $ Is this conjecture true?

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

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 $dz=-2\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to
$$\begin{align*}
\mathfrak I &=\int_0^\infty \frac1{1+\left(e^{2x}\frac{1-x}{1+x}\right)^2}\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}\\
&=\int_1^{-\infty} \frac1{1+z^2}\frac{-\mathrm dz}{2}\\
&=\frac12\int_{-\infty}^1 \frac{\mathrm d z}{1+z^2}\\
&=\frac12\left[\arctan{z}\right]_{-\infty}^1
\end{align*}$$

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

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