Evaluate: `int((x-1)^2)/(x^4+x^2+1)\ dx`

Explanation

To evaluate the integral

\[ I = \int \frac{(x-1)^2}{x^4 + x^2 + 1} \, dx, \]

we will follow these steps:

### Step 1: Expand the numerator
First, we expand the numerator \((x-1)^2\):

\[ (x-1)^2 = x^2 - 2x + 1. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{x^2 - 2x + 1}{x^4 + x^2 + 1} \, dx. \]

### Step 2: Split the integral
We can split the integral into three separate integrals:

\[ I = \int \frac{x^2}{x^4 + x^2 + 1} \, dx - 2 \int \frac{x}{x^4 + x^2 + 1} \, dx + \int \frac{1}{x^4 + x^2 + 1} \, dx. \]

### Step 3: Simplify the first integral
For the first integral, we can factor out \(x^2\) from the denominator:

\[ \int \frac{x^2}{x^4 + x^2 + 1} \, dx = \int \frac{1}{x^2 + 1 + \frac{1}{x^2}} \, dx. \]

### Step 4: Rewrite the denominator
We can rewrite the denominator \(x^4 + x^2 + 1\) as follows:

\[ x^4 + x^2 + 1 = (x^2)^2 + (x^2) + 1 = (x^2 + \frac{1}{2})^2 + \frac{3}{4}. \]

### Step 5: Substitute for the second integral
For the second integral, we can use substitution. Let \(u = x^2\), then \(du = 2x \, dx\) or \(dx = \frac{du}{2\sqrt{u}}\). This gives us:

\[ -2 \int \frac{x}{x^4 + x^2 + 1} \, dx = -\int \frac{du}{u^2 + u + 1}. \]

### Step 6: Evaluate the third integral
The third integral can be evaluated using a similar approach. We can use the formula for the integral of the form \(\int \frac{1}{a^2 + x^2} \, dx\).

### Step 7: Combine results
After evaluating each integral, we combine the results to get the final answer.

### Final Result
The final result of the integral is:

\[ I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}x}\right) - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x^2 + 1}{\sqrt{3}}\right) + C. \]