Explain how to solve this problem---Question Stem:
Let I(x) = $\int \frac{(x+1)}{x(1+xe^x)^2} dx, x > 0$. If $\lim_{x \to \infty} I(x) = 0$, then I(1) is equal to :
Options:
A: $\frac{e+1}{e+2} - \log_e (e+1)$
B: $\frac{e+1}{e+2} + \log_e (e+1)$
C: $\frac{e+2}{e+1} - \log_e (e+1)$
D: $\frac{e+2}{e+1} + \log_e (e+1)$
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We have an integral problem. We need to evaluate the indefinite integral I(x) equals the integral of x plus 1 over x times 1 plus x e to the x squared, dx. Given that the limit of I(x) as x approaches infinity equals zero, we need to find I(1). Let's start by solving this step by step using substitution method.
Let's use substitution to solve this integral. We set u equals 1 plus x e to the x. Taking the derivative, du equals e to the x times 1 plus x dx. We can rewrite x plus 1 dx as du over e to the x. From our substitution, x e to the x equals u minus 1, so e to the x equals u minus 1 over x. Substituting these into our integral, we get the integral of 1 over u squared times u minus 1 du.
Now we perform partial fraction decomposition on 1 over u squared times u minus 1. We write this as A over u plus B over u squared plus C over u minus 1. Multiplying both sides by u squared times u minus 1, we get 1 equals A u times u minus 1 plus B times u minus 1 plus C u squared. Setting u equals 0, we get B equals negative 1. Setting u equals 1, we get C equals 1. Comparing coefficients of u squared, we get A equals negative 1. Therefore, our partial fraction decomposition is negative 1 over u minus 1 over u squared plus 1 over u minus 1.
We integrate each term of our partial fractions. The integral of negative 1 over u minus 1 over u squared plus 1 over u minus 1 gives us negative natural log of u plus 1 over u plus natural log of u minus 1 plus a constant C zero. This simplifies to natural log of u minus 1 over u plus 1 over u plus C zero. Substituting back u equals 1 plus x e to the x, we get I of x equals natural log of x e to the x over 1 plus x e to the x plus 1 over 1 plus x e to the x plus C zero. Using the limit condition that the limit of I of x as x approaches infinity equals zero, we find that C zero equals zero. Therefore, our final expression is I of x equals natural log of x e to the x over 1 plus x e to the x plus 1 over 1 plus x e to the x.
Now we evaluate I of 1 by substituting x equals 1 into our expression. I of 1 equals natural log of e over 1 plus e plus 1 over 1 plus e. Using logarithm properties, natural log of e over 1 plus e equals natural log of e minus natural log of 1 plus e, which equals 1 minus natural log of 1 plus e. So I of 1 equals 1 minus natural log of 1 plus e plus 1 over 1 plus e. Combining the terms, we get 1 plus e plus 1 over 1 plus e minus natural log of 1 plus e, which simplifies to e plus 2 over e plus 1 minus log base e of e plus 1. This matches option C.