Question

\(\int \frac{(x+1)\,dx}{x(1+xe^x)} = \ldots \ldots\)

• A. \(log\frac{xe^x}{1+xe^x}+c\)
• B. \(log\frac{1+xe^x}{xe^x}+c\)
• C. \(log\frac{(x+1)e^x}{xe^x}+c\)
• D. \(log\frac{(x+1)e^x}{1+xe^x}+c\)

Hint:

Multiply the top and bottom by e^x so that the numerator is the derivative of x e^x.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Notice that \(\frac{d}{dx}(xe^x) = (x + 1)e^x\). This suggests the substitution \(t = xe^x\).

Step 2: Key Formula or Approach:
Multiply numerator and denominator by \(e^x\): \(\frac{(x+1)e^x}{xe^x(1 + xe^x)}\).

Step 3: Detailed Explanation:
With \(t = xe^x\), \(dt = (x + 1)e^x\,dx\), so the integral is \(\int\frac{dt}{t(1 + t)}\).
Partial fractions: \(\frac{1}{t(1+t)} = \frac1t - \frac{1}{1+t}\).
\[ \int\left(\frac1t - \frac1{1+t}\right)dt = \log t - \log(1 + t) + c = \log\frac{xe^x}{1 + xe^x} + c \]

Final Answer:
The integral is \(\log\frac{xe^x}{1 + xe^x} + c\), option (A). \[ \boxed{\log\frac{xe^x}{1+xe^x}+c} \]