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

Question

77547 views around the world

You can reuse this answer
Creative Commons License

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

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

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

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

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

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

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

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

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