Question

\(\int \frac{(x+1)}{x(1+xe^x)^2}dx =\)

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

Hint:

Put t = x e^x; then (x+1)/x dx becomes dt/t and partial fractions finish it.

Answer 1

The Correct Option is B

Step 1: Pick the substitution:
The bracket has \(1 + xe^x\), so put \(t = xe^x\). Then \(dt = (e^x + xe^x)dx = e^x(x+1)dx\). Dividing by \(t = xe^x\) gives \(\frac{dt}{t} = \frac{x+1}{x}\,dx\).

Step 2: Rewrite the integral:
The integrand is \(\frac{1}{(1+xe^x)^2}\cdot\frac{x+1}{x}\,dx\). It becomes
\[ \int \frac{dt}{t(1+t)^2} \]

Step 3: Use partial fractions:
Write \(\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}\). Putting \(t = 0\) gives \(A = 1\). Putting \(t = -1\) gives \(C = -1\). Matching the \(t^2\) coefficient gives \(A + B = 0\), so \(B = -1\). Therefore
\[ \frac{1}{t(1+t)^2} = \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \]

Step 4: Integrate and substitute back:

\[ \ln t - \ln(1+t) + \frac{1}{1+t} + c = \ln\left(\frac{xe^x}{1+xe^x}\right) + \frac{1}{1+xe^x} + c \]

Step 5: Why the other options are wrong:
The log term is \(+\ln\frac{t}{1+t}\), so options (A) and (D) with a leading minus are wrong. The last term is \(+\frac{1}{1+t}\), since \(\int -\frac{dt}{(1+t)^2} = +\frac{1}{1+t}\), so option (D) fails again. Option (C) has \((1+xe^x)\) added, which would come from a wrong integral of the squared term.

Final Answer:
The integral equals \(\ln\left(\frac{xe^x}{1+xe^x}\right) + \frac{1}{1+xe^x} + c\), option (B). \[ \boxed{\log\left(\frac{xe^x}{1+xe^x}\right)+\frac{1}{1+xe^x}+c} \]