Q108 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics

$\displaystyle\int \dfrac{x + 1}{x(1 + x e^x)} dx =$

  • A
    $\log\left|\dfrac{c(1 + x e^x)}{x e^x}\right|$
  • Correct answer
    B
    $\log\left|\dfrac{c x e^x}{1 + x e^x}\right|$
  • C
    $\log\left|\dfrac{c}{x e^x(1 + x e^x)}\right|$
  • D
    $\log|c x e^x(1 + x e^x)|$

Solution

Let $I = \displaystyle\int \dfrac{x + 1}{x(1 + x e^x)} dx$

Multiplying the numerator and the denominator by $e^x$:

$I = \displaystyle\int \dfrac{(x + 1)e^x}{x e^x(1 + x e^x)} dx$

Let $x e^x = t$. Differentiating both sides with respect to $x$:

$(x e^x + e^x) dx = dt \Rightarrow (x + 1)e^x dx = dt$

Substituting $t$ and $dt$ into the integral:

$I = \displaystyle\int \dfrac{dt}{t(1 + t)}$

Using partial fractions:

$I = \displaystyle\int \left( \dfrac{1}{t} - \dfrac{1}{1 + t} \right) dt$

Integrating with respect to $t$:

$I = \log|t| - \log|1 + t| + C$

$I = \log\left|\dfrac{t}{1 + t}\right| + \log|c|$

$I = \log\left|\dfrac{c t}{1 + t}\right|$

Substituting back $t = x e^x$:

$I = \log\left|\dfrac{c x e^x}{1 + x e^x}\right|$

Answer: $\log\left|\dfrac{c x e^x}{1 + x e^x}\right|$

Hence, the correct answer is option: 'B'.

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