Qual é a integral de # x / (1 + x ^ 2) #?

Deixei #u(x)=1+x^2" "# então #" "du(x) =2xdx #
#" "#
#color(blue)((d(u(x)))/2=xdx)#
#" "#
Comece a resolver a integral.
#" "#
#intx/(x^2+1)dx#
#" "#
#=intcolor(blue)((d(u(x)))/(2u(x))#
#" "#
#=1/2int(du(x))/(u(x))#
#" "#
#=1/2lnabs(u(x))+C#
#" "#
#=1/2lnabs(x^2+1)+C#
#" "#
Porque #x^2+1>0 " " then " " abs(x^2+1)=x^2+1#
#" " #
Portanto,
#" "#
#intx/(x^2+1)dx=1/2ln(x^2+1)+C#