JEE MainMathematicsEasyDefinite Integration

$\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2x}{1 + x^{2}}$
dx is equal to

Show answer and solution

I = \int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2x}{1 + x^{2}}dx

x = \tan\theta

dx = \sec^{2}\theta d\theta

I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\frac{\mathcal{l}n2tan\theta}{1 + \tan^{2}\theta}\sec^{2}\theta d\theta

I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n2tan\theta d\theta

.......(1)

I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n2\cot\theta d\theta

.......(2)
(1) + (2)

2I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n4d\theta

I = \frac{\pi}{4}\mathcal{l}n4

I = \frac{\pi}{2}\mathcal{l}n2

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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