Key Concepts and Formulas

• Integration by Substitution: If $\int f(g(x))g'(x) dx$ exists, then substituting $u = g(x)$ and $du = g'(x) dx$ transforms the integral to $\int f(u) du$.
• Partial Fraction Decomposition: A rational function can be decomposed into simpler fractions. For example, $\frac{1}{P(x)Q(x)} = \frac{A}{P(x)} + \frac{B}{Q(x)}$ under certain conditions.
• Limits at Infinity: Understanding how functions behave as $x$ approaches infinity is crucial for evaluating definite integrals and constants of integration.

Step-by-Step Solution

Step 1: Simplify the integral using substitution

We are given the integral $I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x$ Let $1+xe^x = t$. Then, differentiating both sides with respect to $x$, we have $e^x + xe^x = \frac{dt}{dx}$ $e^x(1+x)dx = dt$ $(x+1)dx = e^{-x}dt$ Substituting into the integral, we get $I(x) = \int \frac{e^{-x}}{x t^2} dt$ Since $t = 1 + xe^x$, we have $xe^x = t-1$, which implies $e^x = \frac{t-1}{x}$. Thus, $e^{-x} = \frac{x}{t-1}$. Substituting this back into the integral gives us: $I(x) = \int \frac{x}{t-1} \cdot \frac{1}{xt^2} dt = \int \frac{1}{(t-1)t^2} dt$

Step 2: Decompose the integrand using partial fractions

We decompose the integrand $\frac{1}{(t-1)t^2}$ using partial fractions. Let $\frac{1}{(t-1)t^2} = \frac{A}{t-1} + \frac{B}{t} + \frac{C}{t^2}$ Multiplying both sides by $(t-1)t^2$, we get $1 = A t^2 + B t(t-1) + C(t-1)$ $1 = A t^2 + B t^2 - Bt + Ct - C$ $1 = (A+B)t^2 + (C-B)t - C$ Comparing coefficients, we have the following system of equations: $A+B = 0$ $C-B = 0$ $-C = 1$ From the third equation, $C = -1$. Then, from the second equation, $B = C = -1$. Finally, from the first equation, $A = -B = 1$. Therefore, $\frac{1}{(t-1)t^2} = \frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}$

Step 3: Integrate the decomposed expression

Now, we integrate term by term: $I(x) = \int \left(\frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}\right) dt = \int \frac{1}{t-1} dt - \int \frac{1}{t} dt - \int \frac{1}{t^2} dt$ $I(x) = \ln|t-1| - \ln|t| + \frac{1}{t} + C$

Step 4: Substitute back for t

Substituting $t = 1 + xe^x$, we have $I(x) = \ln|xe^x| - \ln|1+xe^x| + \frac{1}{1+xe^x} + C = \ln\left|\frac{xe^x}{1+xe^x}\right| + \frac{1}{1+xe^x} + C$

Step 5: Evaluate the constant of integration using the given limit

We are given that $\lim_{x \to \infty} I(x) = 0$. Thus, $0 = \lim_{x \to \infty} \left(\ln\left|\frac{xe^x}{1+xe^x}\right| + \frac{1}{1+xe^x} + C\right)$ We can rewrite the fraction inside the logarithm as: $\frac{xe^x}{1+xe^x} = \frac{e^x}{\frac{1}{x}+e^x}$ As $x \to \infty$, $\frac{1}{x} \to 0$, so $\lim_{x \to \infty} \frac{e^x}{\frac{1}{x}+e^x} = \lim_{x \to \infty} \frac{e^x}{e^x} = 1$. Also, $\lim_{x \to \infty} \frac{1}{1+xe^x} = 0$. Therefore, $0 = \ln|1| + 0 + C$ $0 = 0 + 0 + C$ Thus, $C = 0$.

Step 6: Find I(1)

We have $I(x) = \ln\left|\frac{xe^x}{1+xe^x}\right| + \frac{1}{1+xe^x}$. We want to find $I(1)$: $I(1) = \ln\left|\frac{1 \cdot e^1}{1+1 \cdot e^1}\right| + \frac{1}{1+1 \cdot e^1} = \ln\left(\frac{e}{1+e}\right) + \frac{1}{1+e}$ $I(1) = \ln(e) - \ln(1+e) + \frac{1}{1+e} = 1 - \ln(1+e) + \frac{1}{1+e}$ $I(1) = \frac{1+e}{1+e} + \frac{1}{1+e} - \ln(1+e) = \frac{2+e}{1+e} - \ln(1+e)$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when performing partial fraction decomposition.
• Limit Evaluation: When evaluating limits involving exponentials, remember that $e^x$ grows much faster than any polynomial as $x$ goes to infinity.
• Constant of Integration: Always remember to include the constant of integration, and use the given conditions to solve for it.

Summary

We solved the indefinite integral by using substitution and partial fraction decomposition. After substituting back to the original variable, we used the given limit condition to determine the constant of integration. Finally, we evaluated the integral at $x=1$ to obtain the final answer.

Final Answer

The final answer is \boxed{\frac{e+2}{e+1}-\log _{e}(e+1)}, which corresponds to option (C). However, the "Correct Answer" provided is (A) $\frac{e+1}{e+2}-\log _{e}(e+1)$. Let's check the algebra again:

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

It seems there was an error in copying the options. We have arrived at $\frac{e+2}{e+1} - \ln(e+1)$. However, the correct option is claimed to be (A) $\frac{e+1}{e+2}-\log _{e}(e+1)$. Let's verify if $\frac{e+2}{e+1} = \frac{e+1}{e+2}$. This is not true.

Since we are confident in our derivation, it is likely there is a typo in the provided options or the given "Correct Answer". If the correct answer is indeed (A), there is an error in the problem statement or the question itself, as our calculations show the answer should be $\frac{e+2}{e+1}-\log_e(e+1)$.

Given the constraint to match the answer, let's examine the provided solution again. There are no apparent errors in the steps. The final answer is indeed $\frac{e+2}{e+1} - \ln(e+1)$.

The question is flawed. The correct answer is $\frac{e+2}{e+1}-\log _{e}(e+1)$, which does not match any of the options.

If we were forced to choose an option that is closest, we would choose (A) because of the $-\log_e(e+1)$ term. However, the fraction is incorrect.

Let us assume there is a typo in the question and that $I(x) = \int \frac{x+1}{x(1+xe^x)^2} dx$. Then $I(1) = \frac{e+2}{e+1} - \log(e+1)$ None of the answer choices match.

The final answer is $\frac{e+2}{e+1}-\log _{e}(e+1)$, which is not among the options. There is likely an error in the question or the options.

Considering the given correct answer (A) is incorrect, and our derived solution leads to $\frac{e+2}{e+1}-\log_e(e+1)$, there is either an error in the options or the provided correct answer. The final answer is \boxed{\frac{e+2}{e+1}-\log _{e}(e+1)}. However, this does not correspond to any of the options.