Question

Let \(I(x)=\int\frac{x+1}{x(1+xe^x)^2} dx\), x>0. If \(\lim\limits_{x\rightarrow\infin}I(x)=0\), then I(1) is equal to

• A. \(\frac{e+2}{e+1}-log_e(e+1)\)
• B. \(\frac{e+2}{e+1}+log_e(e+1)\)
• C. \(\frac{e+1}{e+2}-log_e(e+1)\)
• D. \(\frac{e+1}{e+2}+log_e(e+1)\)

Hint:

When integrating complex expressions, use substitution to simplify the integrand step-by-step.

Answer 1

The Correct Option is A

Step 1: Substitute \(1 + x e^x = t\).
\[ \text{Let } 1 + x e^x = t \implies (x e^x + e^x) \, dx = dt. \] \[ (x + 1) \, dx = \frac{dt}{e^x} = \frac{dt}{t - 1}. \] Step 2: Simplify the integral.
\[ I(x) = \int \frac{dt}{t^2(t-1)} = \int \left[\frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}\right] \, dt. \] \[ I(x) = \ln|t-1| - \ln|t| + \frac{1}{t}. \] Step 3: Evaluate \(I(1)\).
\[ I(1) = \ln\left(\frac{e}{1+e}\right) + \frac{1}{1+e}. \] \[ I(1) = \ln(e) - \ln(1+e) + \frac{1}{1+e} = 1 - \ln(1+e) + \frac{1}{1+e}. \] Final Answer: \(I(1) = \frac{e+2}{e+1} - \log(e+1)\).