Question

If \(\int \frac{x+1}{x(1+xe^{x})^{2}}dx=\log |\frac{xe^{x}}{1+xe^{x}}|+f(x)+c\), then \(f(x)=\)

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

Hint:

Multiply by e^x and substitute t = x e^x, then use partial fractions.

Answer 1

The Correct Option is A

Step 1: Verify by Differentiation
Differentiate $\log\left|\dfrac{t}{1+t}\right|+\dfrac{1}{1+t}$ with respect to $t$.

Step 2: Compute
$\dfrac{1}{t}-\dfrac{1}{1+t}-\dfrac{1}{(1+t)^{2}}=\dfrac{(1+t)^{2}-t(1+t)-t}{t(1+t)^{2}}=\dfrac{1}{t(1+t)^{2}}$.

Step 3: Conclude
This equals the transformed integrand, so the extra term $f(x)$ is $\dfrac{1}{1+xe^{x}}$.

Final Answer:
The extra term is 1/(1 + x e^x). \[ \boxed{\text{(A) }\dfrac{1}{1+xe^{x}}} \]