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

Options

A. `sqrt((x^(4)+x^(2)+1)/(x))+c`
B. `sqrt(x^(4)+2-(1)/(x^(2)))+c`
C. `sqrt(x^(2)+(1)/(x^(2))+1)+c`
D. `sqrt((x^(4)-x^(2)+1)/(x))+c`

Answer

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

### Step 1: Simplify the Integral
We start by rewriting the integral:
\[ \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx = \int \frac{(x^4 - 1)}{x^2} \cdot \frac{1}{\sqrt{x^4 + x^2 + 1}} \, dx \]
This can be simplified to:
\[ \int \frac{x^2 - \frac{1}{x^2}}{\sqrt{x^4 + x^2 + 1}} \, dx \]

### Step 2: Factor Out Common Terms
Next, we can factor out \( x^2 \) from the numerator:
\[ = \int \frac{x^2 \left(1 - \frac{1}{x^4}\right)}{\sqrt{x^4 + x^2 + 1}} \, dx \]

### Step 3: Substitute
Now, we will use a substitution. Let:
\[ t = x^2 + 1 + \frac{1}{x^2} \]
Then, we differentiate \( t \):
\[ dt = \left(2x - \frac{2}{x^3}\right) dx \]
This gives us:
\[ dx = \frac{dt}{2 \left(x - \frac{1}{x^3}\right)} \]

### Step 4: Rewrite the Integral
Substituting \( t \) into the integral, we adjust for the factor of 2:
\[ \int \frac{x^2 \left(1 - \frac{1}{x^4}\right)}{\sqrt{t}} \cdot \frac{dt}{2 \left(x - \frac{1}{x^3}\right)} \]

### Step 5: Simplify Further
After substituting and simplifying, we find:
\[ \int \frac{1}{\sqrt{t}} \, dt \]

### Step 6: Integrate
The integral of \( \frac{1}{\sqrt{t}} \) is:
\[ 2\sqrt{t} + C \]

### Step 7: Substitute Back
Now we substitute back \( t \):
\[ = 2\sqrt{x^2 + 1 + \frac{1}{x^2}} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx = 2\sqrt{x^2 + 1 + \frac{1}{x^2}} + C \]

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