To solve the integral \(\int \frac{x^2 + 1}{x^4 - 2x^2 + 1} \, dx\), we can follow these steps:
### Step 1: Simplify the Denominator
The denominator \(x^4 - 2x^2 + 1\) can be factored. Notice that it can be rewritten as:
\[
x^4 - 2x^2 + 1 = (x^2 - 1)^2
\]
### Step 2: Rewrite the Integral
Now, we can rewrite the integral as:
\[
\int \frac{x^2 + 1}{(x^2 - 1)^2} \, dx
\]
### Step 3: Manipulate the Numerator
Next, we can manipulate the numerator \(x^2 + 1\) by adding and subtracting \(2x\):
\[
x^2 + 1 = (x^2 - 2x + 2x + 1) = (x^2 - 2x) + (2x + 1)
\]
Thus, we can rewrite the integral as:
\[
\int \frac{(x^2 - 2x) + (2x + 1)}{(x^2 - 1)^2} \, dx
\]
### Step 4: Separate the Integral
Now, we can separate the integral into two parts:
\[
\int \frac{x^2 - 2x}{(x^2 - 1)^2} \, dx + \int \frac{2x + 1}{(x^2 - 1)^2} \, dx
\]
### Step 5: Factor the Denominator
Recognize that \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\). Thus, we can rewrite the integrals:
\[
\int \frac{x^2 - 2x}{(x - 1)^2 (x + 1)^2} \, dx + \int \frac{2x + 1}{(x - 1)^2 (x + 1)^2} \, dx
\]
### Step 6: Simplify the First Integral
For the first integral:
\[
\int \frac{x^2 - 2x}{(x^2 - 1)^2} \, dx = \int \frac{(x - 1)^2}{(x^2 - 1)^2} \, dx
\]
This simplifies to:
\[
\int \frac{1}{x + 1} \, dx
\]
### Step 7: Solve the Integrals
Now, we can solve each integral:
1. For \(\int \frac{1}{x + 1} \, dx\):
\[
= \ln |x + 1| + C_1
\]
2. For \(\int \frac{2x + 1}{(x^2 - 1)^2} \, dx\):
We can use substitution \(u = x^2 - 1\), \(du = 2x \, dx\):
\[
= \int \frac{1}{u^2} \, du = -\frac{1}{u} + C_2 = -\frac{1}{x^2 - 1} + C_2
\]
### Step 8: Combine the Results
Combining the results from both integrals, we get:
\[
\int \frac{x^2 + 1}{(x^2 - 1)^2} \, dx = \ln |x + 1| - \frac{1}{x^2 - 1} + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \frac{x^2 + 1}{(x^2 - 1)^2} \, dx = \ln |x + 1| - \frac{1}{x^2 - 1} + C
\]
