Question:

If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

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The trick of multiplying by \( e^{-x} \) turns the denominator into a sum whose derivative is in the numerator.
Alternatively, solve using \( 1 = (e^x + 1) - e^x \) in the numerator.
Updated On: Sep 11, 2026
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Solution and Explanation

Concept:
• Finding a function from its derivative involves integration.
• \( F(x) = \int F'(x) dx + C \).
• Use substitution for logarithmic integrals.

Step 1:
Set up the integral for F(x)
\[ F(x) = \int \frac{1}{e^x + 1} dx \]
Multiply numerator and denominator by \( e^{-x} \):
\[ F(x) = \int \frac{e^{-x}}{1 + e^{-x}} dx \]

Step 2:
Perform substitution
Let \( 1 + e^{-x} = u \), then \( -e^{-x} dx = du \implies e^{-x} dx = -du \).
\[ F(x) = \int -\frac{1}{u} du = -\log |u| + C \]
\[ F(x) = -\log(1 + e^{-x}) + C \]

Step 3:
Use the initial condition to find C
Given \( F(0) = \log(1/2) \).
\[ -\log(1 + e^0) + C = \log(1/2) \implies -\log 2 + C = \log(1/2) \]
\[ -\log 2 + C = -\log 2 \implies C = 0 \]

Step 4:
Write the final function
\[ F(x) = -\log(1 + e^{-x}) = \log\left(\frac{1}{1 + 1/e^x}\right) = \log\left(\frac{e^x}{e^x + 1}\right) \]
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