To solve the integral

$I = \int \frac{2x}{x^2 - 4x + 5} \, dx,$

we can approach this by using a substitution or completing the square for the quadratic expression in the denominator.

Step 1: Complete the square for the quadratic expression in the denominator

We start with $x^2 - 4x + 5$. Completing the square:

$x^2 - 4x + 5 = (x - 2)^2 + 1.$

So the integral becomes:

$I = \int \frac{2x}{(x - 2)^2 + 1} \, dx.$

Step 2: Substitution

Now, let’s make a substitution to simplify the integral further. Set

$u = x - 2,$

which implies

$du = dx \quad \text{and} \quad x = u + 2.$

Substituting into the integral:

$I = \int \frac{2(u + 2)}{u^2 + 1} \, du.$

Step 3: Split the integral

Now, split the integral into two parts:

$I = \int \frac{2u}{u^2 + 1} \, du + \int \frac{4}{u^2 + 1} \, du.$

The first part, $\int \frac{2u}{u^2 + 1} \, du$, is a standard integral:

$\int \frac{2u}{u^2 + 1} \, du = \ln(u^2 + 1).$

The second part, $\int \frac{4}{u^2 + 1} \, du$, is a standard arctangent integral:

$\int \frac{4}{u^2 + 1} \, du = 4 \tan^{-1}(u).$

Step 4: Combine the results

Now, combine both integrals:

$I = \ln(u^2 + 1) + 4 \tan^{-1}(u) + C.$

Step 5: Substitute back $u = x - 2$

Finally, substitute back $u = x - 2$ into the result:

$I = \ln((x - 2)^2 + 1) + 4 \tan^{-1}(x - 2) + C.$

Thus, the solution to the integral is:

$\boxed{ \ln((x - 2)^2 + 1) + 4 \tan^{-1}(x - 2) + C.}$

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Would you like a further breakdown of any of the steps or a different approach to the problem? Here are some related questions you might find helpful:

1. How do you handle integrals involving quadratic expressions in the denominator?
2. What is the relationship between the arctangent and the inverse tangent function?
3. What is the derivative of the arctangent function?
4. How can completing the square help simplify integrals?
5. What happens if the constant term inside the square is negative?

Tip: Completing the square is often a powerful technique for integrals with quadratic denominators, especially when they resemble the form of $\frac{1}{x^2 + a^2}$, which leads directly to an arctangent function.