If $f(x)=\underset{0}{\overset{x}{\int <!--\; \int \; -->}}g(t)\ln <!--\; \; -->\left(\frac{1-<!--\; -\; -->t}{1+t}\right)dt$ and g is odd continuous function and $\underset{-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{\overset{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{\int <!--\; \int \; -->}}(f(x)+\frac{x^2\cos <!--\; \; -->x}{(1+e^x)})dx=\frac{{\mathrm{\pi <!--\; \pi \; -->}}^2}{{\mathrm{\alpha <!--\; \alpha \; -->}}^2}-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}$ then α is

The correct answer is 2.

Let us solve the problem step-by-step.

Step 1: Determine the symmetry (parity) of the function $f\left(x\right)$
We are given the function defined as:
$f\left(x\right)={\int }_0^{x}g\left(t\right)\ln \left(\frac{1-t}{1+t}\right)dt$
where $g\left(t\right)$ is an odd continuous function. This means $g(-t)=-g\left(t\right)$.
Let us evaluate $f(-x)$ using the substitution $t=-y$, so that $dt=-dy$:
$f(-x)={\int }_0^{-x}g\left(t\right)\ln \left(\frac{1-t}{1+t}\right)dt$
$={\int }_0^{x}g(-y)\ln \left(\frac{1+y}{1-y}\right)(-dy)$
Since $g(-y)=-g\left(y\right)$ and $\ln \left(\frac{1+y}{1-y}\right)=-\ln \left(\frac{1-y}{1+y}\right)$, we have:
$f(-x)=-{\int }_0^x(-g(y\left)\right)(-\ln (\frac{1-y}{1+y}\left)\right)dy=-{\int }_0^{x}g\left(y\right)\ln \left(\frac{1-y}{1+y}\right)dy=-f\left(x\right)$
Therefore, $f\left(x\right)$ is an odd function.

Step 2: Evaluate the given definite integral
Let the integral be:
$I={\int }_{-\pi /2}^{\pi /2}\left(f\right(x)+\frac{x^2\cos x}{1+e^x})dx$
Since the interval of integration $[-\pi /2,\pi /2]$ is symmetric, and $f\left(x\right)$ is an odd function, its integral over this interval is zero:
${\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx=0$
Thus, the integral simplifies to:
$I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+e^x}dx$

Step 3: Simplify using the property of definite integrals
Using the property ${\int }_a^{b}h\left(x\right)dx={\int }_a^{b}h(a+b-x)dx$, where $a=-\pi /2$ and $b=\pi /2$ (so $a+b=0$), we get:
$I={\int }_{-\pi /2}^{\pi /2}\frac{(-x{)}^2\cos (-x)}{1+e^{-x}}dx={\int }_{-\pi /2}^{\pi /2}\frac{x^2{e}^x\cos x}{1+e^x}dx$
Adding the two representations of $I$:
$2I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x(1+e^x)}{1+e^x}dx={\int }_{-\pi /2}^{\pi /2}{x}^2\cos xdx$
Since $x^2\cos x$ is an even function, we can write:
$2I=2{\int }_0^{\pi /2}{x}^2\cos xdx\Rightarrow I={\int }_0^{\pi /2}{x}^2\cos xdx$

Step 4: Integration by parts
Integrating by parts twice:
Let $u=x^2,dv=\cos xdx\Rightarrow du=2xdx,v=\sin x$:
$I={\left[x^2\sin x\right]}_0^{\pi /2}-2{\int }_0^{\pi /2}x\sin xdx$
$={\left(\frac{\pi }{2}\right)}^2\left(1\right)-0-2{[-x\cos x+\sin x]}_0^{\pi /2}$
$=\frac{{\pi }^2}{4}-2\left[\right(-\frac{\pi }{2}\cdot 0+1)-(0+0\left)\right]$
$=\frac{{\pi }^2}{4}-2$

Step 5: Compare to the given form
We are given that:
$I=\frac{{\pi }^2}{{\alpha }^2}-\alpha$
Comparing this with our calculated value $I=\frac{{\pi }^2}{4}-2$, we see:
${\alpha }^2=4\Rightarrow \alpha =2$
which also satisfies $\alpha =2$ in the second term. Thus, $\alpha =2$.