How to evaluate the following integral $\int \frac{1}{x^2-x+1} dx$?

Question

$\int \frac{1}{x^2-x+1} dx$ my attempts : $\int \frac{1}{x^2-x+1} dx = \int \frac{1}{(x-\frac{1}{2})^2+ \frac{3}{4}} dx$
let $ x - \frac{1}{2} = t => x = t + \frac{1}{2} $ and $dx=dt$
we get $\int \frac{4}{4t^2 + 3}dt $ someone could help to evaluate it or help with something that work ? thank in advance

Answer 1

Try this

$$\int\dfrac{x^2}{x^2-x+1} \ dx=\int\dfrac{x-1}{x^2-x+1}+1\ dx$$ $$=\int\dfrac{x-1}{x^2-x+1} \ dx+\int1\ dx$$

Now compute $\int\dfrac{x-1}{x^2-x+1}\ dx$

$$\int\dfrac{x-1}{x^2-x+1}\ =\dfrac{x-1}{\left(x-\dfrac12\right)^2+\dfrac34}\ dx$$ Apply $u$ substitution $u=x-\dfrac12$ and so on....

Edit:

If you want to compute $\int\dfrac{4}{4t^2+3}\ dt$

Take $t=\dfrac{\sqrt{3}}{2}v$ and you get $\int\dfrac{4}{2\sqrt{3}(v^2+1)}\ dv\implies\dfrac{2}{\sqrt{3}}\int\dfrac{1}{v^2+1}\ dv$ and you can easily continue from here.

Answer 2

You can easily continue from here with few steps; $\int \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}$

$\int \frac{1}{(x-\frac{1}{2})^2+(\frac{\sqrt3}{2})^2}$

Now the equation forms an identity of $\int \frac {1}{x^2+a^2}$ which equal to $\frac{1}{a}\arctan\frac{x}{a}$

after applying it we get; $\frac{2}{\sqrt3}arctan\frac{(x-\frac{1}{2}) }{\frac{\sqrt3}{2}} $
simplify it, and we will easily get the answer.

Answer 3

Try $$x=\frac{1-t}{1+t}$$ so the integral reduces to $$-\int{\frac{2}{1+3{{t}^{2}}}}dt$$