Let f(x) = ∫ _ 0 ^ x g(t) _ e ((1 - t)/(1 + t))dt , where g is a...

Given: $f\left(x\right) = {∫}_{0}^{x}g\left(t\right)\log \left(\dfrac{1 - t}{1 + t}\right)dt$

$⇒ f\left(- x\right) = {∫}_{0}^{- x}g$\left(t\right)log$\left(\dfrac{1 - t}{1 + t}\right)dt$

$⇒ f\left(- x\right) = - {∫}_{0}^{x}g\left(- y\right)\log \left(\dfrac{1 + y}{1 - y}\right)dy$

$= - {∫}_{0}^{x}g ( y ) \ln \left(\dfrac{1 - y}{1 + y}\right)dy$ ($g$ is odd)

$f ( - x ) = - f ( x ) ⇒ f$ is also odd.

Now,

$I = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\left(f ( x ) + \dfrac{{x}^{2}\cos x}{1 + {e}^{x}}\right)dx . . . \left(i\right)$

$⇒ I = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\left(f ( - x ) + \dfrac{{x}^{2}{e}^{x}\cos x}{1 + {e}^{x}}\right)dx . . . \left(ii\right)$

$⇒ 2I = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}{x}^{2}cosxdx$

$⇒ 2I = 2{∫}_{0}^{\dfrac{\pi}{2}}{x}^{2}cosxdx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}{x}^{2}cosxdx$

$⇒ I = {\left({x}^{2}\sin x\right)}_{0}^{\dfrac{\pi}{2}} - {∫}_{0}^{\dfrac{\pi}{2}}2xsinxdx$

$⇒ I = \dfrac{{\pi}^{2}}{4} - 2{\left(- xcosx + \sin x\right)}_{0}^{\dfrac{\pi}{2}}$

$⇒ I = \dfrac{{\pi}^{2}}{4} - 2 ( 0 + 1 )$

$⇒ I = \dfrac{{\pi}^{2}}{4} - 2$

$⇒ I = {\left(\dfrac{\pi}{2}\right)}^{2} - 2$

$∴ \alpha = 2$