Let $$I(x) = \int \dfrac{x+1}{x(1+xe^x)^2} dx$$, $$x > 0$$. If $$\lim_{x \to \infty} I(x) = 0$$ then $$I(1)$$ is equal to

The integral to be evaluated is $$I(x) = \int \dfrac{x+1}{x(1+xe^x)^2} dx$$ for $$x>0$$, subject to the condition that $$\lim_{x \to \infty} I(x) = 0$$. We then seek the value of $$I(1)$$.

Introducing the substitution $$u = 1 + xe^x$$ leads to $$du = e^x(1+x)\,dx$$, so that $$(1+x)\,dx = \frac{du}{e^x}$$. Since $$xe^x = u-1$$, the integral becomes
$$I = \int \frac{1}{x\,u^2}\,\frac{du}{e^x} = \int \frac{1}{xe^x\,u^2}\,du = \int \frac{1}{(u-1)\,u^2}\,du.$$

Decomposing the integrand into partial fractions gives
$$\frac{1}{(u-1)u^2}=\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^2}.$$
Integration then yields
$$I = \ln|u-1| - \ln|u| + \frac{1}{u} + C.$$

Reverting to the original variable via $$u = 1 + xe^x$$ transforms this expression into
$$I(x) = \ln(xe^x) - \ln(1+xe^x) + \frac{1}{1+xe^x} + C = x + \ln x - \ln(1+xe^x) + \frac{1}{1+xe^x} + C.$$

To determine the constant, observe that as $$x\to\infty$$ one has $$\ln(1+xe^x)\approx x+\ln x$$ and $$\frac{1}{1+xe^x}\to0$$, so the expression for $$I(x)$$ tends to $$C$$. Imposing the condition $$\lim_{x\to\infty}I(x)=0$$ then forces $$C=0$$.

Finally, substituting $$x=1$$ gives
$$I(1)=1+0-\ln(1+e)+\frac{1}{1+e}=\frac{1+e+1}{1+e}-\ln(e+1)=\frac{e+2}{e+1}-\log_e(e+1).$$

The correct answer is Option 1: $$\frac{e+2}{e+1}-\log_e(e+1).$$