Let I x = ∫ x + 1 x 1 + x e x 2 d x ,   x > 0 . If lim x → ∞ I x = 0 then I 1 is…

Given,

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

Now let $1+x{e}^x=t$

$\Rightarrow e^x\left(x+1\right)dx=dt$

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

$\Rightarrow I\left(x\right)=\int \frac{\left(1-t^2\right)+t^2}{\left(t-1\right)t^2}dt$

$\Rightarrow I\left(x\right)=\int \frac{-\left(t+1\right)}{t^2}+\frac{1}{t-1}dt$

$\Rightarrow I\left(x\right)=\int -\frac{1}{t}-\frac{1}{t^2}+\frac{1}{t-1}dt$

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

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

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

$\Rightarrow \underset{x\to \infty }{\lim }I\left(x\right)=\underset{x\to \infty }{\lim }\left[\ln \left(1-\frac{1}{x{e}^x+1}\right)+\frac{1}{x{e}^x+1}+C\right]$

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

$\Rightarrow C=0$

Now finding,

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

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