What is the integral of #x/(1+x^2)#?

1 Answer
Nov 21, 2016

Answer:

#intx/(x^2+1)dx=1/2ln(x^2+1)+C#

Explanation:

Let #u(x)=1+x^2" "# then #" "du(x) =2xdx #
#" "#
#color(blue)((d(u(x)))/2=xdx)#
#" "#
Start solving the 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#
#" "#
Because #x^2+1>0 " " then " " abs(x^2+1)=x^2+1#
#" " #
Therefore,
#" "#
#intx/(x^2+1)dx=1/2ln(x^2+1)+C#