`int((x^(2)-1)dx)/(xsqrt(1+3x^(2)+x^(4))).`

Answer

To solve the integral

\[ \int \frac{x^2 - 1}{x \sqrt{1 + 3x^2 + x^4}} \, dx, \]

we will follow a series of steps to simplify and evaluate the integral.

### Step 1: Simplify the Denominator
First, we notice that the expression under the square root in the denominator can be rewritten. We have:

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

Thus, we can rewrite the integral as:

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

### Step 2: Split the Fraction
Next, we can split the fraction:

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

So, we can rewrite the integral as:

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

### Step 3: Integrate Each Term
Now we can integrate each term separately:

1. The first term:

\[ \int \frac{1}{x} \, dx = \ln |x| + C_1. \]

2. The second term can be integrated using partial fractions:

\[ \frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}. \]

Multiplying through by the denominator \(x(x^2 + 1)\) gives:

\[ 1 = A(x^2 + 1) + (Bx + C)x. \]

Expanding and equating coefficients, we find:

- For \(x^2\): \(A + B = 0\)
- For \(x\): \(C = 0\)
- For constant: \(A = 1\)

From \(A + B = 0\), we have \(B = -1\). Thus, we can write:

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

So, we integrate:

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

The second integral can be solved using substitution \(u = x^2 + 1\), \(du = 2x \, dx\):

\[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln |x^2 + 1| + C_2. \]

### Step 4: Combine Results
Combining all the results, we have:

\[ \int \frac{x^2 - 1}{x \sqrt{1 + 3x^2 + x^4}} \, dx = \ln |x| - \left( \ln |x| - \frac{1}{2} \ln |x^2 + 1| \right) + C. \]

This simplifies to:

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

### Final Answer
Thus, the final answer is:

\[ \int \frac{x^2 - 1}{x \sqrt{1 + 3x^2 + x^4}} \, dx = \frac{1}{2} \ln |x^2 + 1| + C. \]