# Evaluate the indefinite integral. `int dx/(x^2-x+1)`

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### 1 Answer

`int (dx)/(x^2-x+1)`

To start, express the denominator of integrand in the form `a^2+u^2`. To do so, group the first two terms and apply the completing the square method.

`int (dx)/((x^2-x)+1)=int (dx)/((x^2-x+1/4)+1-1/4)`

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

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

Then, apply u-substitution method. Let,

`u=x-1/2`

Differentiate u.

`du=dx`

Replace the x variable of the integrand with u.

`int (du)/(3/4+u^2)`

To evaluate, use the integral formula `int (du)/(a^2+u^2)=1/a tan^-1u/a+C` .

`int (du)/(3/4+u^2) = int (du)/((sqrt3/2)^2 +u^2) = 1/(sqrt3/2)tan^(-1)(x-1/2)/(sqrt3/2)+ C `

`= 2/sqrt3 tan^(-1) (2(x-1/2))/sqrt3+ C= 2/sqrt3 tan^(-1) (2x-1)/sqrt3`

Then, rationalize the denominator.

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

**Hence, `int (dx)/(x^2-x+1)=(2sqrt3)/3 tan^(-1) (sqrt3(2x-1))/3 +C` .**