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

Options

A. `sqrt((x^4+x^2+1)/(x))+C`
B. `(x^2)/(sqrt(x^4+x^2+1))+C`
C. `x(x^4+x^2+1)^(3//2)+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 integrand
We can rewrite the integrand by separating the terms in the numerator:

\[ \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} = \frac{x^4}{x^2 \sqrt{x^4 + x^2 + 1}} - \frac{1}{x^2 \sqrt{x^4 + x^2 + 1}}. \]

This simplifies to:

\[ \frac{x^2}{\sqrt{x^4 + x^2 + 1}} - \frac{1}{x^2 \sqrt{x^4 + x^2 + 1}}. \]

### Step 2: Make a substitution
Let us make the substitution:

\[ t = x^4 + x^2 + 1. \]

Then, we differentiate \(t\):

\[ dt = (4x^3 + 2x) \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3 + 2x}. \]

### Step 3: Express \(x^2\) in terms of \(t\)
From our substitution, we can express \(x^2\) in terms of \(t\):

\[ x^2 = \sqrt{t - 1 - x^2}. \]

### Step 4: Substitute back into the integral
Now we substitute \(x^2\) and \(dx\) back into the integral. The integral becomes:

\[ \int \frac{x^2}{\sqrt{t}} \cdot \frac{dt}{4x^3 + 2x} - \int \frac{1}{x^2 \sqrt{t}} \cdot \frac{dt}{4x^3 + 2x}. \]

### Step 5: Simplify and integrate
After substituting and simplifying, we can integrate:

\[ \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} + C. \]

### Step 6: Substitute back for \(t\)
Finally, we substitute back \(t = x^4 + x^2 + 1\):

\[ = 2\sqrt{x^4 + x^2 + 1} + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \sqrt{x^4 + x^2 + 1} + C. \]