`int(x^2-1)/(x^4+x^2+1)`dx is equal to

Options

A. `1/2log((x^2+x+1)/(x^2-x+1))+C`
B. `1/2log((x^2-x-1)/(x^2+x+1))+C`
C. `log((x^2-x+1)/(x^2+x+1))+C`
D. `1/2log((x^2-x+1)/(x^2+x+1))+C`

Answer

To solve the integral \( \int \frac{x^2 - 1}{x^4 + x^2 + 1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the given integral:
\[ I = \int \frac{x^2 - 1}{x^4 + x^2 + 1} \, dx \]

### Step 2: Factor the Denominator
Notice that the denominator can be rewritten:
\[ x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1) \]
However, for our purpose, we will not factor it just yet.

### Step 3: Split the Numerator
We can separate the integral into two parts:
\[ I = \int \frac{x^2}{x^4 + x^2 + 1} \, dx - \int \frac{1}{x^4 + x^2 + 1} \, dx \]

### Step 4: Simplify the First Integral
For the first integral, we can rewrite:
\[ \int \frac{x^2}{x^4 + x^2 + 1} \, dx = \int \frac{1}{x^2 + 1 + \frac{1}{x^2}} \, dx \]
Let \( u = x + \frac{1}{x} \). Then, \( du = (1 - \frac{1}{x^2}) \, dx \).

### Step 5: Change of Variables
Using the substitution \( t = x + \frac{1}{x} \):
\[ dx = \frac{1}{1 - \frac{1}{x^2}} \, dt = \frac{1}{1 - \frac{1}{(t^2 - 2)}} \, dt \]

### Step 6: Rewrite the Integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int \frac{1}{t^2 - 1} \, dt \]

### Step 7: Solve the Integral
The integral \( \int \frac{1}{t^2 - 1} \, dt \) can be solved using partial fractions:
\[ \frac{1}{t^2 - 1} = \frac{1}{(t-1)(t+1)} = \frac{1/2}{t-1} - \frac{1/2}{t+1} \]
Thus,
\[ I = \frac{1}{2} \ln |t - 1| - \frac{1}{2} \ln |t + 1| + C \]

### Step 8: Substitute Back
Substituting back for \( t \):
\[ I = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C \]
And since \( t = x + \frac{1}{x} \):
\[ I = \frac{1}{2} \ln \left| \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} \right| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x^2 - 1}{x^4 + x^2 + 1} \, dx = \frac{1}{2} \ln \left| \frac{x^2 - x + 1}{x^2 + x + 1} \right| + C \]