The integral given in the image is:

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

To solve this, we first simplify the expression by rewriting the denominator:

$x^2 - 2x = (x - 1)^2 - 1$

So the integral becomes:

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

However, this still doesn't simplify directly, so we should try another approach. Let's perform polynomial long division first, or recognize that we can split the fraction:

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

Now, the integral becomes:

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

The first part is straightforward:

$\int 1 \, dx = x$

For the second part, let's simplify the integral by completing the square for the denominator:

$x^2 - 2x = (x - 1)^2 - 1$

Thus, the integral becomes:

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

Now, make the substitution $u = x - 1$, hence $du = dx$ and $2x + 1 = 2(u + 1) + 1 = 2u + 3$. The integral now is:

$\int \frac{2u + 3}{u^2 - 1} \, du$

This integral can be split into two separate integrals:

$\int \frac{2u}{u^2 - 1} \, du + \int \frac{3}{u^2 - 1} \, du$

The first integral is simplified by recognizing that the numerator is the derivative of the denominator:

$\int \frac{2u}{u^2 - 1} \, du = \ln|u^2 - 1|$

The second integral is a standard integral that can be recognized as the sum of partial fractions:

$\frac{1}{u^2 - 1} = \frac{1}{2} \left(\frac{1}{u - 1} - \frac{1}{u + 1}\right)$

Therefore, the integral becomes:

$\int \frac{3}{u^2 - 1} \, du = \frac{3}{2} \ln \left| \frac{u - 1}{u + 1} \right|$

Now, substituting back $u = x - 1$:

$\int \frac{2x + 1}{x^2 - 2x} \, dx = \ln|x^2 - 2x - 1| + \frac{3}{2} \ln \left| \frac{x - 2}{x} \right|$

Finally, combining all parts together:

$\int \frac{x^2 + 1}{x^2 - 2x} \, dx = x + \ln|x^2 - 2x - 1| + \frac{3}{2} \ln \left| \frac{x - 2}{x} \right| + C$

This is the integral of the given expression. Would you like more details or have any questions?

Here are five related questions for practice:

1. How would you approach integrating $\int \frac{x + 2}{x^2 - 4} \, dx$?
2. What is the integral of $\int \frac{2x + 1}{x^2 + x} \, dx$?
3. How do partial fractions help in integrating rational functions?
4. What is the difference between substitution and polynomial long division when integrating?
5. Can you derive the integral of $\int \frac{1}{x^2 - 1} \, dx$?

Tip: When integrating rational functions, always consider simplifying the expression first with polynomial division or substitution to make the integral more straightforward.