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

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Answer 1 · 2018-07-12T18:07:01

#(2sqrt(3))/3 arctan( (2x+1)/sqrt(3) ) + c #

#x^2 + x+1 = (x+1/2)^2 + 3/4 #

#=> int 1/((x+1/2)^2 +3/4) dx #

let #3/4 u^2 = (x+1/2)^2 #

#=> sqrt3 /2 u = x+1/2 #

#=> sqrt(3)/2 du = dx #

#=> sqrt(3)/2 int 1/(3/4 u^2 + 3/4) du #

#=> (2sqrt(3))/3 int 1/(u^2+1) du #

#=> (2sqrt(3))/3 arctan(u) + c #

# = (2sqrt(3))/3 arctan( (2x+1)/sqrt(3) ) + "constant" #

Answer 2 · 2018-07-12T18:10:51

# 2/sqrt3*arctan{(2x+1)/sqrt3}+C#.

#"Prerequsite :"int1/{(x+a)^2+b^2}dx=1/b*arctan((x+a)/b)+c#.

#int1/(x^2+x+1)dx#,

#=int1/{(x^2+x+1/4)+3/4}dx..............."[completing square]"#,

#=int1/{(x+1/2)^2+(sqrt3/2)^2}dx#,

#=1/(sqrt3/2)*arctan{(x+1/2)/(sqrt3/2)}#,

#=2/sqrt3*arctan{(2x+1)/sqrt3}+C#.