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### Answers (1)

Answer:

$\pi \log 2$

Hint:

To solve this equation we suppose x as tan x.

Given:

$\int_{0}^{\infty} \log \left(x+\frac{1}{x}\right)\left(\frac{1}{1+x^{2}}\right) d x$

Solution:

Let

\begin{aligned} &I=\int_{0}^{\infty} \log \left(x+\frac{1}{x}\right)\left(\frac{1}{1+x^{2}}\right) d x \\\\ &x=\tan x \end{aligned}

\begin{aligned} &d x=\sec ^{2} \theta d \theta \\\\ &\text { When } x=0, \tan \theta=0, \theta=0 \\\\ &\text { When } x=\infty, \tan \theta=\infty, \theta=\frac{\pi}{2} \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log (\tan \theta+\cot \theta) \cdot \frac{\sec ^{2} \theta d \theta}{1+\tan ^{2} \theta} \\\\ &=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} \log 1-\log (\sin \theta \cos \theta) d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \log \sin \theta d \theta-\int_{0}^{\frac{\pi}{2}} \log \cos \theta d \theta \\\\ &=-\left(\frac{\pi}{2} \log 2\right)-\left(-\frac{\pi}{2} \log 2\right) \end{aligned}

$I=0$

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