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test
123 days ago by Alexey_Kalmykov
integrate(1/(1+x^2),x); show(integrate(1/(1+x^2),x,-1,1))
arctan(x)
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, \pi
arctan(x)
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, \pi