The value of integral ₀^{∞} {1}{1+e^{x}} d x is...

Correct Answer is : log⁡2 2log2

Correct Answer is : log⁡2 2log2

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MHTCET 19 Apr 2026 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

\int_{0}^{\infty} \frac{1}{1+e^{x}} d x

is...

Solution:

Concept:

Use the substitution

t = e^{-x}

to convert the integral into a standard logarithmic form.

Explanation:

Let

t = e^{-x}

, so

dx = -\frac{dt}{t}

Then

\frac{1}{1+e^x} = \frac{t}{1+t}

When

x=0

t=1

; and as

x \to \infty

t \to 0

Substituting into the integral:

\int_{0}^{\infty} \frac{1}{1+e^x}\,dx = \int_{1}^{0} \frac{t}{1+t}\left(-\frac{dt}{t}\right) = \int_{0}^{1} \frac{dt}{1+t}.

Evaluating:

[\log(1+t)]_{0}^{1} = \log 2 - \log 1 = \log 2

Answer:

Option A:

\log 2

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