Let I(x) = ∫ x + 1 x (1 + x e ^ x ) ^ 2 dx, x > 0 . If _ x → ∞ I(x)...

Given,

$I\left(x\right) = ∫ \dfrac{x + 1}{x{\left(1 + x{e}^{x}\right)}^{2}}dx$

Now let $1 + x{e}^{x} = t$

$⇒ {e}^{x}\left(x + 1\right)dx = dt$

So, $I\left(x\right) = ∫ \dfrac{1}{\left(t - 1\right){t}^{2}}dt$

$⇒ I\left(x\right) = ∫ \dfrac{\left(1 - {t}^{2}\right) + {t}^{2}}{\left(t - 1\right){t}^{2}}dt$

$⇒ I\left(x\right) = ∫ \dfrac{- \left(t + 1\right)}{{t}^{2}} + \dfrac{1}{t - 1}dt$

$⇒ I\left(x\right) = ∫ - \dfrac{1}{t} - \dfrac{1}{{t}^{2}} + \dfrac{1}{t - 1}dt$

$⇒ I\left(x\right) = - \ln t + \dfrac{1}{t} + \ln \left(t - 1\right) + C$

$⇒ I\left(x\right) = \ln \left(\dfrac{x{e}^{x}}{x{e}^{x} + 1}\right) + \dfrac{1}{x{e}^{x} + 1} + C$

Also given,$\lim_{x \to \infty}I\left(x\right) = 0$

$⇒ \lim_{x \to \infty}I\left(x\right) = \lim_{x \to \infty}\left[\ln \left(1 - \dfrac{1}{x{e}^{x} + 1}\right) + \dfrac{1}{x{e}^{x} + 1} + C\right]$

$⇒ \lim_{x \to \infty}I\left(x\right) = \left[\ln \left(1 - 0\right) + 0 + C\right]$

$⇒ C = 0$

Now finding,

$I\left(1\right) = \ln \left(\dfrac{e}{e + 1}\right) + \dfrac{1}{e + 1} = 1 + \dfrac{1}{e + 1} - \ln ( e + 1 )$

$⇒ I\left(1\right) = \dfrac{e + 2}{e + 1} - \ln \left(e + 1\right)$