To find $u$ in this substitution step, the idea is to simplify the expression within the integral by using a substitution that involves completing the square.

Step-by-Step Solution:

1. Original Integral: The given integral is: $\int \frac{1}{x^2 - x + 1} \, dx$

2. Complete the Square: The expression in the denominator, $x^2 - x + 1$, can be completed as follows: $x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$ This simplifies the denominator, as we can now rewrite the integral as: $\int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}} \, dx$

3. Substitution: Let $u = \frac{2x - 1}{\sqrt{3}}$. This substitution was chosen because: $\left(x - \frac{1}{2}\right)^2 + \frac{3}{4} = \frac{u^2}{4} + \frac{3}{4} = \frac{u^2 + 1}{4}$ Using this substitution, we find: $\frac{d}{dx} \left(\frac{2x - 1}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}$ so $du = \frac{2}{\sqrt{3}} \, dx$, which gives $dx = \frac{\sqrt{3}}{2} \, du$.

4. Rewrite the Integral: Substitute $u$ and $dx = \frac{\sqrt{3}}{2} \, du$ into the integral: $\int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}} \, dx = \int \frac{1}{\frac{u^2 + 1}{4}} \cdot \frac{\sqrt{3}}{2} \, du$ Simplifying this gives: $\frac{2}{\sqrt{3}} \int \frac{1}{u^2 + 1} \, du$

5. Result: Now we have the integral in terms of $u$: $\frac{2}{\sqrt{3}} \int \frac{1}{u^2 + 1} \, du$

Would you like a detailed solution for the remaining integration?

Additional Questions to Explore

1. How do we choose a substitution for integrals involving quadratics?
2. What are other methods for integrating rational functions?
3. Why is completing the square helpful in integration?
4. How does this integral relate to trigonometric substitution?
5. Can partial fraction decomposition apply here?

Tip: Always verify the derivative of your substitution matches the integral's differential before proceeding.