I tried to compute the integral by using the substitution $x=e^y$ but I couldn't find an answer (I checked it converges by checking limits at both ends by using the comparison test).
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Split the integral into the intervals $(0, 1)$ and $(1, +\infty)$. Hence,
\begin{align*} \int_{0}^{\infty} \frac{\ln x}{1+x^2} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln x}{1+x^2} \, \mathrm{d}x + \int_{1}^{\infty} \frac{\ln x}{1+x^2} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\ln x}{1+x^2} + \int_{0}^{1} \frac{\ln \frac{1}{x}}{1+\left ( \frac{1}{x} \right )^2} \frac{1}{x^2} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\ln x}{1+x^2} \, \mathrm{d}x - \int_{0}^{1} \frac{\ln x}{1+x^2} \, \mathrm{d}x \\ &= 0 \end{align*}
and this is your answer.
Tolaso
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You were thinking about a correct substitution. Indeed: $$\int_0^{\infty}\frac{\ln x}{1+x^2}=\int_{-\infty}^{\infty}\frac{y e^y dy}{1+e^{2y}}=\int_{-\infty}^{\infty}\frac{y dy}{e^{-y}+e^{y}}=0.$$
stokes-line
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Thank you! But why is the last part equal to 0 ? – Lakhdar Mohamed Amine Mar 22 '20 at 12:12
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1The integrated function is odd (changes sign under $y\to-y$). – stokes-line Mar 22 '20 at 12:12