Definite Integration
Definite + Exponential
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^1\frac{e^x}{1+e^x}\,dx\) [JEE Main 2019]</p>
<li>\(\ln(1+e)-\ln 2\)</li>
<li>\(\ln(e+1)\)</li>
<li>\(\ln\dfrac{e}{1+e}\)</li>
<li>\(1-\ln 2\)</li>
Step-by-Step Solution
Key Concept: d/dx[ln(1+eˣ)] = eˣ/(1+eˣ). So \int = [ln(1+eˣ)]_0^1 = ln(1+e) - ln(2).
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<p>$$\int_0^1\frac{e^x}{1+e^x}dx = [\ln(1+e^x)]_0^1 = \ln(1+e)-\ln(1+1)=\ln(1+e)-\ln 2$$</p>
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Correct Answer: A