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

Answer

To evaluate the integral

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

we will follow a systematic approach.

### Step 1: Simplify the Integral
First, we can rewrite the integral by factoring out \(x^2\) from both the numerator and the denominator:

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

This simplifies to:

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

### Step 2: Rewrite the Denominator
Next, we can rewrite the denominator \(x^4 + x^2 + 1\) as follows:

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

This allows us to express the integral in a more manageable form.

### Step 3: Substitute
Now, we can use the substitution \(t = x + \frac{1}{x}\). Then, we differentiate:

\[ dt = \left(1 - \frac{1}{x^2}\right) dx. \]

Thus, we can rewrite the integral in terms of \(t\):

\[ \int \frac{1 - \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \, dx = \int \frac{dt}{\text{(expression in terms of t)}}. \]

### Step 4: Evaluate the Integral
The integral can be evaluated using the formula for the integral of the form:

\[ \int \frac{1}{t^2 - a^2} \, dt = \frac{1}{2a} \ln \left| \frac{t - a}{t + a} \right| + C. \]

In our case, \(a = 1\). So, we have:

\[ \int \frac{1}{t^2 - 1} \, dt = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C. \]

### Step 5: Back Substitute
Now we back substitute \(t = x + \frac{1}{x}\):

\[ \int \frac{x^2 - 1}{x^4 + x^2 + 1} \, dx = \frac{1}{2} \ln \left| \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} \right| + C. \]

### Final Result
Thus, the final result for the integral is:

\[ \frac{1}{2} \ln \left| \frac{x^2 - x + 1}{x^2 + x + 1} \right| + C. \]