Question

How do you integrate this? ∫ dx(x²-x+1) I'm stuck on this part (image uploaded)

Answer 1

`=> (2sqrt3)/3 tan^(-1) ( (2x-1) / sqrt3 ) + c`

Explanation:

Carrying on...

Let `3/4 u^2 = (x-1/2)^2`

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

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

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

`=> sqrt3 / 2 int 1 / ( 3/4 ( u^2+1 ) ) du`

`=> (2sqrt3) / 3 int 1/(u^2+1) du`

Using an antiderivative what should be committed to memory...

`=> (2sqrt3)/3 tan^(-1) u +c`

`=> u = (2x-1 )/sqrt3`

`=> (2sqrt3)/3 tan^(-1) ( (2x-1) / sqrt3 ) + c`

Answer 2

This is a tricky little integral, and the solution will not appear obvious at first. Since this is a fraction, we might try to consider using the partial fractions technique, but a quick analysis reveals that this is not possible since `x^2-x+1` is not factorable.

We will try to get this integral to a form that we can actually integrate. Notice the similarity between `int1/(x^2-x+1)dx` and `int1/(x^2+1)dx`; we know that the latter integral evaluates to `arctanx+C`. We will therefore try to get `x^2-x+1` in the form `k(x-a)^2+1`, and then apply the `arctanx` rule.

We will need to complete the square on `x^2-x+1`:
`x^2-x+1`
`=x^2-x+1/4+1-1/4`
`=(x-1/2)^2+3/4`
`=(x-1/2)^2+(sqrt(3)/2)^2`
`=(sqrt(3)/2)^2((x-1/2)^2/(sqrt(3)/2)^2+1)`
`=(sqrt(3)/2)^2(((x-1/2)/(sqrt(3)/2))^2+1)`
(very messy, I know)

Now that we have it in our desired form, we may proceed as follows:
`int1/(x^2-x+1)dx=int1/((sqrt(3)/2)^2(((x-1/2)/(sqrt(3)/2))^2+1))dx`
`=4/3int1/(((x-1/2)/(sqrt(3)/2))^2+1)dx`
`=4/3int1/(((2x-1)/(sqrt(3)))^2+1)dx`
`=4/3*(sqrt(3)/2arctan((2x-1)/sqrt(3)))+C`
`=(2arctan((2x-1)/sqrt(3)))/sqrt(3)+C`