Let $$f(x) = \int_0^x g(t)\log_e\frac{1-t}{1+t}dt$$, where g is a continuous odd function. If $$\int_{-\pi/2}^{\pi/2} \left(f(x) + \frac{x^2\cos x}{1 + e^x}\right) dx = \left(\frac{\pi}{\alpha}\right)^2 - \alpha$$, then $$\alpha$$ is equal to _____.

We are given $$f(x) = \int_0^x g(t)\,\log_e\!\frac{1-t}{1+t}\,dt$$ where $$g$$ is a continuous odd function.

Showing $$f$$ is odd. Substitute $$t = -u$$, $$dt = -du$$ in $$f(-x)$$:

$$f(-x) = \int_0^{-x} g(t)\log_e\!\frac{1-t}{1+t}\,dt = -\int_0^{x} g(-u)\log_e\!\frac{1+u}{1-u}\,du.$$

Since $$g(-u) = -g(u)$$ (odd) and $$\log_e\frac{1+u}{1-u} = -\log_e\frac{1-u}{1+u}$$:

$$f(-x) = -\int_0^x g(u)\log_e\!\frac{1-u}{1+u}\,du = -f(x).$$

So $$f$$ is odd and $$\int_{-\pi/2}^{\pi/2} f(x)\,dx = 0$$.

Evaluating $$\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+e^x}\,dx$$. The function $$h(x) = x^2\cos x$$ satisfies $$h(-x) = h(x)$$, so it is even. For any continuous even $$h$$ we have the identity:

$$\int_{-a}^{a}\frac{h(x)}{1+e^x}\,dx = \int_0^a h(x)\,dx.$$

To prove this, split at 0 and substitute $$x = -u$$ on $$[-a,0]$$:

$$\int_{-a}^{0}\frac{h(x)}{1+e^x}\,dx = \int_0^{a}\frac{h(u)\,e^u}{1+e^u}\,du.$$

Adding the integral on $$[0,a]$$: $$\int_0^a h(u)\!\left(\frac{e^u+1}{1+e^u}\right)du = \int_0^a h(u)\,du$$.

Therefore the integral reduces to $$\int_0^{\pi/2} x^2\cos x\,dx$$.

Computing $$\int_0^{\pi/2} x^2\cos x\,dx$$. Integrate by parts with $$u = x^2$$, $$dv = \cos x\,dx$$, so $$du = 2x\,dx$$, $$v = \sin x$$:

$$\int_0^{\pi/2} x^2\cos x\,dx = \bigl[x^2\sin x\bigr]_0^{\pi/2} - 2\int_0^{\pi/2} x\sin x\,dx = \frac{\pi^2}{4} - 2\int_0^{\pi/2} x\sin x\,dx.$$

For the remaining integral, set $$u = x$$, $$dv = \sin x\,dx$$, giving $$du = dx$$, $$v = -\cos x$$:

$$\int_0^{\pi/2} x\sin x\,dx = \bigl[-x\cos x\bigr]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = 0 + \bigl[\sin x\bigr]_0^{\pi/2} = 1.$$

Therefore $$\int_0^{\pi/2} x^2\cos x\,dx = \frac{\pi^2}{4} - 2$$.

Finding $$\alpha$$. We need $$\left(\frac{\pi}{\alpha}\right)^2 - \alpha = \frac{\pi^2}{4} - 2$$, i.e., $$\frac{\pi^2}{\alpha^2} - \alpha = \frac{\pi^2}{4} - 2$$. Setting $$\alpha = 2$$:

$$\frac{\pi^2}{4} - 2 = \frac{\pi^2}{4} - 2. \quad\checkmark$$

So, the answer is $$2$$.