Question:
For \(x \gt 0\), if \(f(x) = \displaystyle\int_1^x \frac{\log_e t}{(1 + t)}\,dt\), then \(f(e) + f\left(\frac{1}{e}\right)\) is equal to :
For \(x \gt 0\), if \(f(x) = \displaystyle\int_1^x \frac{\log_e t}{(1 + t)}\,dt\), then \(f(e) + f\left(\frac{1}{e}\right)\) is equal to :
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Substitute \(t \to 1/t\) in \(f(1/e)\) and add to \(f(e)\); the sum collapses to \(\int_1^e \frac{\ln t}{t}\,dt\).
Updated On: Nov 4, 2026
- \(\frac{1}{2}\)
- \(-1\)
- \(0\)
- \(1\)
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The Correct Option is A
Solution and Explanation
\(f(1/e) = \int_1^{1/e}\frac{\ln t}{1+t}dt\). Put \(t = 1/u\): \(dt = -\frac{du}{u^2}\), \(\ln t = -\ln u\), limits \(u: 1 \to e\).
\(f(1/e) = \int_1^e \frac{-\ln u}{1 + 1/u}\cdot\left(-\frac{1}{u^2}\right)du = \int_1^e \frac{\ln u}{u(1+u)}du\).
So \(f(e) + f(1/e) = \int_1^e \ln t\left[\frac{1}{1+t} + \frac{1}{t(1+t)}\right]dt = \int_1^e \frac{\ln t}{t}\,dt\).
\(= \left[\frac{(\ln t)^2}{2}\right]_1^e = \frac12\).
Answer: (1) \(\frac{1}{2}\)
\(f(1/e) = \int_1^e \frac{-\ln u}{1 + 1/u}\cdot\left(-\frac{1}{u^2}\right)du = \int_1^e \frac{\ln u}{u(1+u)}du\).
So \(f(e) + f(1/e) = \int_1^e \ln t\left[\frac{1}{1+t} + \frac{1}{t(1+t)}\right]dt = \int_1^e \frac{\ln t}{t}\,dt\).
\(= \left[\frac{(\ln t)^2}{2}\right]_1^e = \frac12\).
Answer: (1) \(\frac{1}{2}\)
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