Let $$f(x)=\int_\limits0^x g(t) \log _{\mathrm{e}}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}$$, where $g$ is a continuous odd function. If $$\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+\mathrm{e}^x}\right) \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$$, then $\alpha$ is equal to _________.

<p>$f(x)=\int_\limits0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$</p> <p>$f(-x)=\int_\limits0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$</p> <p>$f(-x)=-\int_\limits0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$</p> <p>$=-\int_\limits0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y$ (g is odd)</p> <p>$f(-x)=-f(x) \Rightarrow f$ is also odd</p> <p>Now,</p> <p>$$\begin{aligned} & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x \quad \text{... (1)}\\ & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^2 e^x \cos x}{1+e^x}\right) d x \quad \text{... (2)}\\ & 2 I=\int_\limits{-\pi / 2}^{\pi / 2} x^2 \cos x d x=2 \int_0^{\pi / 2} x^2 \cos x d x \end{aligned}$$</p> <p>$$\begin{aligned} & I=\left(x^2 \sin x\right)_0^{\pi / 2}-\int_\limits0^{\pi / 2} 2 x \sin x d x \\ & =\frac{\pi^2}{4}-2\left(-x \cos x+\int \cos x d x\right)_0^{\pi / 2} \\ & =\frac{\pi^2}{4}-2(0+1)=\frac{\pi^2}{4}-2 \Rightarrow\left(\frac{\pi}{2}\right)^2-2 \\ & \therefore \alpha=2 \end{aligned}$$</p>