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$\int_0^{\infty} \frac{\log \left(1+x^2\right)}{1+x^2} d x=$
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$\pi \log 2$
$\begin{aligned} & \text { Let } I=\int_0^x \frac{\log \left(1+x^2\right)}{1+x^2} d x \\ & \text { Put } x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta\end{aligned}$
$\begin{gathered}\therefore \quad 1=\int_0^{\pi / 2} \log (\sec \theta)^2 d \theta=2 \int_0^{\pi / 2} \log \operatorname{(sed} \theta) \theta \\ =-2 \int_0^{\pi / 2} \log \cos \theta d \theta=-2 \cdot \frac{\pi}{2} \log \frac{1}{2}=-\pi \log \frac{1}{2}=\pi \log 2\end{gathered}$
$\begin{gathered}\therefore \quad 1=\int_0^{\pi / 2} \log (\sec \theta)^2 d \theta=2 \int_0^{\pi / 2} \log \operatorname{(sed} \theta) \theta \\ =-2 \int_0^{\pi / 2} \log \cos \theta d \theta=-2 \cdot \frac{\pi}{2} \log \frac{1}{2}=-\pi \log \frac{1}{2}=\pi \log 2\end{gathered}$
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