Q238 / 270 · NDA 2015 (Phase 2) · Mathematics

$\int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}}$ is equal to where $c$ is the constant of integration

  • A
    $1+\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
  • B
    $\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)+\mathrm{c}$
  • Correct answer
    C
    $\ln \left(1+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$
  • D
    $2 \ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)+\mathrm{c}$

Solution

$\begin{aligned} & \int \frac{\mathrm{dx}}{1+\mathrm{e}^{-\mathrm{x}}} \\ & \Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx} \\ & \text {Let } \mathrm{e}^{\mathrm{x}}+1=\mathrm{t} \\ & \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \\ & =\int \frac{\mathrm{dt}}{\mathrm{t}} \\ & \Rightarrow \log \mathrm{t}+\mathrm{c} \\ & \Rightarrow \log \left(\mathrm{e}^{\mathrm{x}}+1\right)+\mathrm{c}\end{aligned}$

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

Relevant Questions