Q88 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
$\int\dfrac{(x+1)\,dx}{x(1+xe^x)} = \ldots\ldots$
- Correct answerA$\log\left|\dfrac{xe^x}{1+xe^x}\right| + c$
- B$\log\left|\dfrac{1+xe^x}{xe^x}\right| + c$
- C$\log\left|\dfrac{(x+1)e^x}{xe^x}\right| + c$
- D$\log\left|\dfrac{(x+1)e^x}{1+xe^x}\right| + c$
Solution
Let $I = \int\dfrac{x+1}{x(1+xe^x)} dx$
Multiplying the numerator and the denominator by $e^x$:
$I = \int\dfrac{(x+1)e^x}{xe^x(1+xe^x)} dx$
Let $xe^x = t$
Differentiating both sides with respect to $x$:
$(1 \cdot e^x + x \cdot e^x) dx = dt$
$(x+1)e^x dx = dt$
Substituting these into the integral:
$I = \int\dfrac{dt}{t(1+t)}$
Using partial fractions:
$I = \int\left(\dfrac{1}{t} - \dfrac{1}{1+t}\right) dt$
$I = \log|t| - \log|1+t| + c$
$I = \log\left|\dfrac{t}{1+t}\right| + c$
Substituting $t = xe^x$ back into the expression:
$I = \log\left|\dfrac{xe^x}{1+xe^x}\right| + c$
Answer: $\log\left|\dfrac{xe^x}{1+xe^x}\right| + c$
Multiplying the numerator and the denominator by $e^x$:
$I = \int\dfrac{(x+1)e^x}{xe^x(1+xe^x)} dx$
Let $xe^x = t$
Differentiating both sides with respect to $x$:
$(1 \cdot e^x + x \cdot e^x) dx = dt$
$(x+1)e^x dx = dt$
Substituting these into the integral:
$I = \int\dfrac{dt}{t(1+t)}$
Using partial fractions:
$I = \int\left(\dfrac{1}{t} - \dfrac{1}{1+t}\right) dt$
$I = \log|t| - \log|1+t| + c$
$I = \log\left|\dfrac{t}{1+t}\right| + c$
Substituting $t = xe^x$ back into the expression:
$I = \log\left|\dfrac{xe^x}{1+xe^x}\right| + c$
Answer: $\log\left|\dfrac{xe^x}{1+xe^x}\right| + c$
Hence, the correct answer is option: 'A'.