Q90 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics

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

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

Solution

Let $I = \int \dfrac{x+1}{x(1+xe^x)^2} dx$

Multiplying the numerator and the denominator by $e^x$, we get:

$I = \int \dfrac{(x+1)e^x}{xe^x(1+xe^x)^2} dx$

Let $xe^x = t$. Differentiating both sides with respect to $x$, we get:

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

Substituting these into the integral:

$I = \int \dfrac{dt}{t(1+t)^2}$

Using partial fractions:

$\dfrac{1}{t(1+t)^2} = \dfrac{1+t-t}{t(1+t)^2} = \dfrac{1}{t(1+t)} - \dfrac{1}{(1+t)^2}$

$\dfrac{1}{t(1+t)^2} = \dfrac{1+t-t}{t(1+t)} - \dfrac{1}{(1+t)^2} = \dfrac{1}{t} - \dfrac{1}{1+t} - \dfrac{1}{(1+t)^2}$

Integrating with respect to $t$:

$I = \int \left( \dfrac{1}{t} - \dfrac{1}{1+t} - \dfrac{1}{(1+t)^2} \right) dt$

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

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

Substituting back $t = xe^x$:

$I = \log\left(\dfrac{xe^x}{1+xe^x}\right) + \dfrac{1}{1+xe^x} + c$

Answer: $\log\left(\dfrac{xe^x}{1+xe^x}\right) + \dfrac{1}{(1+xe^x)} + c$

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

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