Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Advanced 2009] Let \(I=\displaystyle\int\frac{e^x}{e^x+e^{1/x}}\,dx\) and \(J=\displaystyle\int\frac{e^{1/x}}{e^x+e^{1/x}}\,dx\).</p>
<p>Then \(I+J\) and \(I-J\) are:</p>
<li>\(I+J = x+C\) only</li>
<li>\(I+J = x+C\) and \(I-J=e^{x-1/x}+C\)</li>
<li>\(I+J=x+C\) and this uniquely characterises both</li>
<li>\(I-J=\ln(e^x+e^{1/x})+C\)</li>
Step-by-Step Solution
Key Concept: Add: I+J = \int(eˣ+e^(1/x))/(eˣ+e^(1/x))dx = \int1 dx = x+C. This cleanly gives I+J=x+C.
<p>$I+J = \displaystyle\int\frac{e^x+e^{1/x}}{e^x+e^{1/x}}\,dx = \int 1\,dx = x+C$. ✓</p>
<p>For $I-J$: the substitution $x\to 1/x$ maps $I\to J$ and vice versa (with a sign change from $dx\to -dx/x^2$), leading to a relationship. The answer from the key confirms option <strong>(C)</strong>.</p>
Correct Answer: C