Integration help $\frac{1}{x^2 - x + 1}$?

Question

How do you integrate $\dfrac{1}{x^2 - x + 1}$?

I started with completing the square: $\dfrac{1}{(x-0.5)^2 + 0.75)}$.

Now I'm stuck.

Answer 1

After shifting the variable, you have obtained

$$\int\frac{du}{u^2+\dfrac34},$$ which reminds of the $\arctan$ function, such that

$$\int\frac{dt}{t^2+1}=\arctan t+C.$$

To reduce to this form, you use a scaling of the variable in such a way that the denominator $u^2+\dfrac34$ becomes $\dfrac34t^2+\dfrac34$. This is achieved by setting

$$u=\dfrac{\sqrt3}2t.$$

The rest is yours.

Answer 2

The integral can be solved with a series of substitutions. Starting off where you left off, we have$$\begin{align*}\int\frac {dx}{\left(x-\tfrac 12\right)^2+\tfrac 34} & =\int\frac {du}{u^2+\tfrac 34}\\ & =\frac 43\int\frac {du}{1+\tfrac {4u^2}3}\\ & =\frac 2{\sqrt3}\int\frac {ds}{1+s^2}\\ & =\frac 2{\sqrt3}\arctan s\end{align*}$$