To solve the integral

$\int \frac{1}{x^2 - x + 2} \, dx,$

we start by simplifying the denominator and making it more manageable. The quadratic expression $x^2 - x + 2$ does not factorize easily, so we complete the square.

Step 1: Complete the square for $x^2 - x + 2$

$x^2 - x + 2 = \left(x - \frac{1}{2}\right)^2 + \frac{7}{4}.$

Thus, the integral becomes:

$\int \frac{1}{x^2 - x + 2} \, dx = \int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{7}{4}} \, dx.$

Step 2: Use the standard arctangent formula

The standard integral formula is:

$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C.$

Here, we have a similar structure, but with $\left(x - \frac{1}{2}\right)^2 + \frac{7}{4}$. Let:

$u = x - \frac{1}{2}, \quad \text{so } dx = du.$

Substituting, the integral becomes:

$\int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{7}{4}} \, dx = \int \frac{1}{u^2 + \frac{7}{4}} \, du.$

Rewrite $\frac{7}{4}$ as $\left(\sqrt{\frac{7}{4}}\right)^2 = \left(\frac{\sqrt{7}}{2}\right)^2$. Now the integral becomes:

$\int \frac{1}{u^2 + \left(\frac{\sqrt{7}}{2}\right)^2} \, du.$

Applying the arctangent formula:

$\int \frac{1}{u^2 + \left(\frac{\sqrt{7}}{2}\right)^2} \, du = \frac{2}{\sqrt{7}} \arctan\left(\frac{2u}{\sqrt{7}}\right) + C.$

Step 3: Back-substitute $u = x - \frac{1}{2}$

Substitute $u = x - \frac{1}{2}$ back into the result:

$\int \frac{1}{x^2 - x + 2} \, dx = \frac{2}{\sqrt{7}} \arctan\left(\frac{2\left(x - \frac{1}{2}\right)}{\sqrt{7}}\right) + C.$

Final Answer:

$\boxed{\int \frac{1}{x^2 - x + 2} \, dx = \frac{2}{\sqrt{7}} \arctan\left(\frac{2x - 1}{\sqrt{7}}\right) + C.}$

Would you like further clarification or detailed steps on any part?

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Here are 5 related questions to explore:

1. How would the process change if the quadratic denominator was factorizable?
2. What is the integral of $\frac{1}{x^2 + c^2}$ in general, and how does it differ?
3. How can completing the square help in solving indefinite integrals more broadly?
4. Could this integral be evaluated with a trigonometric substitution instead?
5. How does this result apply in physics or engineering contexts?

Tip: Completing the square is a powerful technique for integrating rational functions and understanding the geometry of quadratic expressions.