Definite Integration
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^1\frac{e^x}{1+e^x}\,dx\) [JEE Main 2019]</p>
\(\ln(1+e)-\ln 2\)
\(\ln(e+1)\)
\(\ln\dfrac{e}{1+e}\)
\(1-\ln 2\)
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).
<div class='solution'>
<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>
Correct Answer: A