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in Definite Integrals by (15 points)
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\( \int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x \)

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Let I = \(\int\limits^{\infty}_{0} \frac{logx}{1+x^2}\,dx\) 

Let x = tanθ 

⇒ dx =  sec2θdθ

\(\therefore I = \int\limits _{0}^{\frac{\pi}2} \frac{log(tan\theta )\,sec^2\theta d\theta}{1+ tan^2\theta}\)

\(= \int\limits _{0}^{\frac{\pi}2} log(tan\theta )d\theta\)   ......(i)    (∵ 1 + tan2θ = sec2θ)

\(\therefore I = \int\limits _{0}^{\frac{\pi}2} log(tan(\frac{\pi}2-\theta))d\theta\)

\(= \int\limits _{0}^{\frac{\pi}2} log(cot \theta )d\theta\)       ......(ii)

On adding (i) & (ii), we got

\(2 I = \int\limits _{0}^{\frac{\pi}2} (log\,tan\theta + log\,cot\theta )d\theta\)

\(= \int\limits _{0}^{\frac{\pi}2} log(tan\theta -cot \theta )d\theta\)

\(= \int\limits _{0}^{\frac{\pi}2} log1d\theta =0\)        (∵ log 1 = 0)

\(\therefore I = 0\)

⇒ \(\int\limits^{\infty}_{0} \frac{logx}{1+x^2}\,dx = 0\)

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