`intx^2sqrt(x^6-1)dx`

Answer

To solve the integral \( \int x^2 \sqrt{x^6 - 1} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int x^2 \sqrt{x^6 - 1} \, dx \]
We can express \( x^6 \) as \( (x^3)^2 \):
\[ I = \int x^2 \sqrt{(x^3)^2 - 1} \, dx \]

### Step 2: Substitution
Let \( t = x^3 \). Then, differentiating both sides gives:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
From our substitution, we can express \( x^2 \, dx \) as:
\[ x^2 \, dx = \frac{dt}{3} \]

### Step 3: Substitute in the integral
Now we substitute \( x^2 \, dx \) and \( \sqrt{x^6 - 1} \):
\[ I = \int \frac{dt}{3} \sqrt{t^2 - 1} \]
We can factor out \( \frac{1}{3} \):
\[ I = \frac{1}{3} \int \sqrt{t^2 - 1} \, dt \]

### Step 4: Use the integral identity
We know the integral:
\[ \int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln |x + \sqrt{x^2 - a^2|} + C \]
In our case, \( a = 1 \):
\[ I = \frac{1}{3} \left( \frac{t}{2} \sqrt{t^2 - 1} - \frac{1}{2} \ln |t + \sqrt{t^2 - 1}| \right) + C \]

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = x^3 \):
\[ I = \frac{1}{3} \left( \frac{x^3}{2} \sqrt{x^6 - 1} - \frac{1}{2} \ln |x^3 + \sqrt{x^6 - 1}| \right) + C \]

### Step 6: Simplify the expression
Distributing \( \frac{1}{3} \):
\[ I = \frac{x^3}{6} \sqrt{x^6 - 1} - \frac{1}{6} \ln |x^3 + \sqrt{x^6 - 1}| + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int x^2 \sqrt{x^6 - 1} \, dx = \frac{x^3}{6} \sqrt{x^6 - 1} - \frac{1}{6} \ln |x^3 + \sqrt{x^6 - 1}| + C \]