Question

Find `int \ 1/(x^2 + x + 1) \ dx`?

Answer 1

`(2/sqrt(3)) arctan((2/sqrt(3))x + 1/(sqrt(3))) + C`

Explanation:

`"Write "x^2+x+1" as "(x+1/2)^2 + 3/4`
`"and this as "(3/4) ((4/3) (x+1/2)^2 + 1)`
`"so we get "(3/4) ((2/sqrt(3)) x + 1/(sqrt(3)))^2 + 1)`
`"so we have"`

`(4/3) (sqrt(3)/2) int (d((2/sqrt(3))x + 1/(sqrt(3)))) / (((2/sqrt(3)) x + 1/(sqrt(3)))^2 + 1)`

`"Now we recognize the integral of "1/(1+x^2)" which is arctan(x)."`

`=> (2/sqrt(3)) arctan((2/sqrt(3))x + 1/(sqrt(3))) + C`

Answer 2

`int \ 1/(x^2 + x + 1) \ dx = 2/sqrt(3) \ arctan( (2x+1)/sqrt(3)) + C`

Explanation:

We seek:

`I = int \ 1/(x^2 + x + 1) \ dx`

First we complete the square on the quadratic denominator:

`I = int \ 1/((x + 1/2)^2-(1/2)^2 + 1) \ dx`
`\ \ = int \ 1/((x + 1/2)^2 + 3/4) \ dx`
`\ \ = int \ 1/((x + 1/2)^2 + (sqrt(3)/2)^2) \ dx`

Now, we can perform a substitution, Let

`u = (2x+1)/sqrt(3) => x = sqrt(3)/2u -1/2`

And differentiating wrt `x` we have:

`(du)/dx =2/sqrt(3)`

If we substitute into the integral, we then get:

`I = int \ 1/((sqrt(3)/2u)^2 + (sqrt(3)/2)^2) \ (sqrt(3)/2) \ du`

`\ \ = sqrt(3)/2 \ int \ 1/((sqrt(3)/2)^2u^2 + (sqrt(3)/2)^2) \ du`

`\ \ = sqrt(3)/2 \ int \ (2/sqrt(3))^2 1/(u^2 +1 ) \ du`

`\ \ = 2/sqrt(3) \ int \ 1/(u^2 +1 ) \ du`

And this is a standard integral, so we can integrate, to get:

`I = 2/sqrt(3) \ arctanu + C`

Then we restore the substitution:

`I = 2/sqrt(3) \ arctan( (2x+1)/sqrt(3)) + C`