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