https://file.aiquickdraw.com/mathbot/file/1724690001062lqaq4ovh.jpg

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.

Learn how to integrate (x^2 + 1) / (x^2 - 2x) with a detailed step-by-step solution. This problem involves integration of rational functions using substitution and partial fractions. Suitable for college-level mathematics.

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