1. Given integral

We need to evaluate

$$I(x)=\int \frac{x+1}{x(1+xe^x)^2}\,dx, \qquad x>0$$

with the condition

$$\lim_{x\to\infty} I(x)=0.$$

Then find $I(1)$.

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2. Look for a useful substitution / derivative pattern

Notice the expression $xe^x$ appears. Let

$$u = xe^x.$$

Then

$$\frac{d}{dx}(xe^x)=e^x(x+1).$$

Also,

$$\frac{x+1}{x}e^{-x} = \frac{d}{dx}(xe^x)\cdot \frac{1}{x e^{2x}}$$

but a cleaner route is to rewrite the integrand as

$$\frac{x+1}{x(1+xe^x)^2} = \frac{(x+1)e^x}{xe^x(1+xe^x)^2}.$$

Since

$$\frac{d}{dx}(xe^x)=(x+1)e^x,$$

we get

$$I(x)=\int \frac{1}{u(1+u)^2}\,du, \qquad \text{where } u=xe^x.$$

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3. Partial fraction decomposition

We decompose

$$\frac{1}{u(1+u)^2}.$$

Assume

$$\frac{1}{u(1+u)^2}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{(1+u)^2}.$$

Then

$$1=A(1+u)^2+Bu(1+u)+Cu.$$

Expanding:

$$1=A(1+2u+u^2)+B(u+u^2)+Cu.$$

So,

$$1=A+(2A+B+C)u+(A+B)u^2.$$

Comparing coefficients:

$$A=1,$$ $$A+B=0 \Rightarrow B=-1,$$ $$2A+B+C=0 \Rightarrow 2-1+C=0 \Rightarrow C=-1.$$

Thus,

$$\frac{1}{u(1+u)^2}=\frac{1}{u}-\frac{1}{1+u}-\frac{1}{(1+u)^2}.$$

Therefore,

$$I(x)=\int\left(\frac{1}{u}-\frac{1}{1+u}-\frac{1}{(1+u)^2}\right)du.$$

Integrating,

$$I(x)=\ln u-\ln(1+u)+\frac{1}{1+u}+C.$$

Substitute back $u=xe^x$:

$$I(x)=\ln(xe^x)-\ln(1+xe^x)+\frac{1}{1+xe^x}+C.$$

Since $x>0$, this is valid.

Also,

$$\ln(xe^x)=\ln x + x.$$

So,

$$I(x)=\ln x + x - \ln(1+xe^x)+\frac{1}{1+xe^x}+C.$$

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4. Use the condition $\lim_{x\to\infty} I(x)=0$ to find $C$

As $x\to\infty$,

$$1+xe^x \sim xe^x.$$

Hence,

$$\ln(1+xe^x) \sim \ln(xe^x)=\ln x + x.$$

Therefore,

$$\ln x + x - \ln(1+xe^x) \to 0,$$

and also

$$\frac{1}{1+xe^x}\to 0.$$

So,

$$\lim_{x\to\infty} I(x)=C.$$

Given this limit is $0$, we get

$$C=0.$$

Thus,

$$I(x)=\ln x + x - \ln(1+xe^x)+\frac{1}{1+xe^x}.$$

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5. Evaluate $I(1)$

Substitute $x=1$:

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

Since $\ln 1=0$,

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

Combine the rational terms:

$$1+\frac{1}{e+1}=\frac{e+1}{e+1}+\frac{1}{e+1}=\frac{e+2}{e+1}.$$

Hence,

$$I(1)=\frac{e+2}{e+1}-\ln(e+1).$$

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6. Match with options

This is exactly

$$\boxed{\frac{e+2}{e+1}-\log_e(e+1)}$$

which corresponds to Option C.

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7. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.