JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of the integral \(\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx\) is:
- A \(\dfrac{\pi\log_e(2)}{2}\)
- B \(\dfrac{\pi\log_e(2)}{4}\)
- C \(1 + \pi\log_e(2)\)
- D \(2 + \pi\log_e(2)\)
Answer & Solution
Correct Answer
(B) \(\dfrac{\pi\log_e(2)}{4}\)
Step-by-step Solution
Detailed explanation
Let \(I = \displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx\) Substitute \(x = 2 \tan \theta\), which gives \(dx = 2 \sec^2 \theta \,d\theta\). The limits of integration change from \(x = 0 \to \theta = 0\) to \(x \to \infty \to \theta = \dfrac{\pi}{2}\).…
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