Question:medium
The value of the integral $\int_0^\infty \frac{\log_e (x)}{x^2 + 4} dx$ is:
The value of the integral $\int_0^\infty \frac{\log_e (x)}{x^2 + 4} dx$ is:
Updated On: Apr 12, 2026
- $\frac{\pi \log_e (2)}{2}$
- $\frac{\pi \log_e (2)}{4}$
- $1 + \pi \log_e (2)$
- $2 + \pi \log_e (2)$
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The Correct Option is B
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