Q92 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
The value of integral $\int_0^\infty \dfrac{1}{1 + e^x}\,dx$ is...
- Correct answerA$\log 2$
- B$\log\left(\dfrac{2}{e}\right)$
- C$-\log e$
- D$\log\left(\dfrac{4}{e}\right)$
Solution
Let $I = \int_0^\infty \dfrac{1}{1 + e^x} dx$
Multiplying the numerator and the denominator by $e^{-x}$:
$I = \int_0^\infty \dfrac{e^{-x}}{e^{-x} + 1} dx$
Let $e^{-x} + 1 = t$
Differentiating both sides, $-e^{-x} dx = dt \Rightarrow e^{-x} dx = -dt$
Limits of integration:
When $x = 0$, $t = e^0 + 1 = 2$
When $x \to \infty$, $t = e^{-\infty} + 1 = 1$
Substituting these into the integral:
$I = \int_2^1 \dfrac{-dt}{t} = \int_1^2 \dfrac{dt}{t}$
$I = [\log t]_1^2 = \log 2 - \log 1 = \log 2$
Answer: $\log 2$
Multiplying the numerator and the denominator by $e^{-x}$:
$I = \int_0^\infty \dfrac{e^{-x}}{e^{-x} + 1} dx$
Let $e^{-x} + 1 = t$
Differentiating both sides, $-e^{-x} dx = dt \Rightarrow e^{-x} dx = -dt$
Limits of integration:
When $x = 0$, $t = e^0 + 1 = 2$
When $x \to \infty$, $t = e^{-\infty} + 1 = 1$
Substituting these into the integral:
$I = \int_2^1 \dfrac{-dt}{t} = \int_1^2 \dfrac{dt}{t}$
$I = [\log t]_1^2 = \log 2 - \log 1 = \log 2$
Answer: $\log 2$
Hence, the correct answer is option: 'A'.