The problem involves evaluating the following indefinite integral:

$\int \frac{x^2 - 4x + 1}{x^2 + 1} \, dx$

Let's break it down step by step.

Step 1: Simplifying the expression

We start by simplifying the integrand. We can divide the numerator by the denominator.

$\frac{x^2 - 4x + 1}{x^2 + 1}$

We will first split the fraction into simpler parts:

$\frac{x^2 - 4x + 1}{x^2 + 1} = \frac{x^2 + 1 - 4x}{x^2 + 1}$

This can be broken into two fractions:

$\frac{x^2}{x^2 + 1} - \frac{4x}{x^2 + 1} + \frac{1}{x^2 + 1}$

Step 2: Integrating each term

Now, we will integrate each term separately.

1. First term:

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

We can rewrite this as:

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

This splits into:

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

The integrals of these are:

$x - \arctan(x)$

2. Second term:

$\int \frac{-4x}{x^2 + 1} \, dx$

This is a standard integral. The integral of $\frac{2x}{x^2 + 1}$ is $\ln(x^2 + 1)$, so the integral of this term is:

$-2 \ln(x^2 + 1)$

3. Third term:

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

This is a standard integral:

$\arctan(x)$

Step 3: Combining everything

Now, putting everything together:

$x - \arctan(x) - 2 \ln(x^2 + 1) + \arctan(x) + C$

The $\arctan(x)$ terms cancel out, so the final answer is:

$x - 2 \ln(x^2 + 1) + C$

Final Answer:

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

Would you like a more detailed explanation or further steps? Let me know if you have any questions!

Here are 5 questions related to this topic:

1. How do you handle similar integrals where the numerator and denominator both contain polynomials?
2. What is the method of partial fraction decomposition, and when is it applicable?
3. Can we use substitution to solve integrals of this type? If so, how?
4. What are some other standard integrals that come in handy for integrals involving rational functions?
5. How do you check the correctness of an indefinite integral?

Tip: Always check if the numerator can be split or simplified before attempting the integral. It makes solving the problem easier!