1. Given

We have

$$f(x)=\int_0^x g(t)\ln\left(\frac{1-t}{1+t}\right)dt,$$

where $g$ is a continuous odd function.

We need to evaluate

$$I=\int_{-\pi/2}^{\pi/2}\left(f(x)+\frac{x^2\cos x}{1+e^x}\right)dx$$

and compare it with

$$I=\left(\frac{\pi}{\alpha}\right)^2-\alpha.$$

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2. Parity of the integrand defining $f$

Let

$$h(t)=g(t)\ln\left(\frac{1-t}{1+t}\right).$$

Since $g$ is odd,

$$g(-t)=-g(t).$$

Also,

$$\ln\left(\frac{1-(-t)}{1+(-t)}\right)=\ln\left(\frac{1+t}{1-t}\right)=-\ln\left(\frac{1-t}{1+t}\right).$$

So the logarithmic factor is also odd. Therefore,

$$h(-t)=g(-t)\ln\left(\frac{1+t}{1-t}\right)=(-g(t))\left(-\ln\left(\frac{1-t}{1+t}\right)\right)=h(t).$$

Hence $h(t)$ is even.

Now

$$f(x)=\int_0^x h(t)\,dt.$$

Integral of an even function from $0$ to $x$ is an odd function. Thus,

$$f(-x)=-f(x).$$

Therefore,

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

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3. Reduce the given integral

So,

$$I=\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+e^x}\,dx.$$

Let

$$J=\int_{-a}^{a}\frac{x^2\cos x}{1+e^x}\,dx,$$

where $a=\pi/2$.

Use the property

$$\int_{-a}^{a} \phi(x)\,dx=\int_{0}^{a}[\phi(x)+\phi(-x)]dx.$$

Here,

$$\phi(x)=\frac{x^2\cos x}{1+e^x}.$$

Then

$$\phi(-x)=\frac{(-x)^2\cos(-x)}{1+e^{-x}}=\frac{x^2\cos x}{1+e^{-x}}.$$

Thus,

$$\phi(x)+\phi(-x)=x^2\cos x\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right).$$

Now,

$$\frac{1}{1+e^{-x}}=\frac{e^x}{1+e^x},$$

so

$$\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=\frac{1}{1+e^x}+\frac{e^x}{1+e^x}=1.$$

Therefore,

$$\phi(x)+\phi(-x)=x^2\cos x.$$

Hence

$$I=\int_0^{\pi/2} x^2\cos x\,dx.$$

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4. Evaluate $\int_0^{\pi/2} x^2\cos x\,dx$

Integrate by parts:

Take

$$u=x^2, \quad dv=\cos x\,dx.$$

Then

$$du=2x\,dx, \quad v=\sin x.$$

So

$$\int x^2\cos x\,dx=x^2\sin x-\int 2x\sin x\,dx.$$

Now evaluate

$$\int 2x\sin x\,dx=2\int x\sin x\,dx.$$

Again by parts, let

$$u=x, \quad dv=\sin x\,dx,$$

then

$$du=dx, \quad v=-\cos x.$$

Thus

$$\int x\sin x\,dx=-x\cos x+\int \cos x\,dx=-x\cos x+\sin x.$$

Therefore,

$$\int x^2\cos x\,dx=x^2\sin x-2(-x\cos x+\sin x)$$ $$=x^2\sin x+2x\cos x-2\sin x.$$

Now apply limits $0$ to $\pi/2$:

$$I=\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\pi/2}.$$

At $x=\pi/2$:

$$\sin\frac{\pi}{2}=1, \quad \cos\frac{\pi}{2}=0,$$

so value is

$$\frac{\pi^2}{4}-2.$$

At $x=0$:

$$0+0-0=0.$$

Hence

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

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5. Compare with the given form

Given

$$I=\left(\frac{\pi}{\alpha}\right)^2-\alpha.$$

So,

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

Matching the coefficients immediately gives

$$\alpha=2,$$

because

$$\left(\frac{\pi}{2}\right)^2-2=\frac{\pi^2}{4}-2.$$

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6. Final answer

$$\boxed{2}$$