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Question: \[\int_{0}^{\infty}\frac{\log(1 + x^{2})}{1 + x^{2}}dx =\]...

∫0∞log⁡(1+x2)1+x2dx=\int_{0}^{\infty}\frac{\log(1 + x^{2})}{1 + x^{2}}dx =

A

πlog⁡12\pi\log\frac{1}{2}

B

πlog⁡2\pi\log 2

C

2πlog⁡122\pi\log\frac{1}{2}

D

2πlog⁡22\pi\log 2

Answer

πlog⁡2\pi\log 2

Explanation

Solution

Let I=∫0∞log⁡(1+x2)1+x2dxI = \int_{0}^{\infty}{\frac{\log(1 + x^{2})}{1 + x^{2}}dx}

Put x=tan⁡θ⇒dx=sec⁡2θdθ,x = \tan\theta \Rightarrow dx = \sec^{2}\theta d\theta,

∴\therefore I=∫0π/2log⁡(sec⁡θ)2dθ=2∫0π/2logsecθdθI = \int_{0}^{\pi/2}{\log(\sec\theta)^{2}d\theta = 2\int_{0}^{\pi/2}{{logsec}\theta d\theta}}

=−2∫0π/2logcosθdθ=−2.π2log⁡12=−πlog⁡12=πlog⁡2= - 2\int_{0}^{\pi/2}{{logcos}\theta d\theta = - 2.\frac{\pi}{2}\log\frac{1}{2}} = - \pi\log\frac{1}{2} = \pi\log 2.