For any natural number m, evaulate,
`int(x^(3m)+x^(2m)+x^(m))(2x^(2m)+3x^9m)+6^(t//m)dx, x gt0`
`int (x^(2)-1)/(x^(3)sqrt(2x^(4)-2x^(2)+1)` dx is equal to:

Answer

To evaluate the integral

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

we will follow a step-by-step approach.

### Step 1: Simplify the Integral

First, we can simplify the expression under the square root. Notice that:

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

This doesn't simplify directly, so we will keep it as is for now.

### Step 2: Rewrite the Integral

Now, we can rewrite the integral as:

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

### Step 3: Split the Integral

We can split the integral into two parts:

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

### Step 4: Use Substitution

Let's use the substitution \( u = 2x^4 - 2x^2 + 1 \). Then, we differentiate:

\[ \frac{du}{dx} = 8x^3 - 4x = 4x(2x^2 - 1) \implies du = (4x(2x^2 - 1)) \, dx. \]

Thus,

\[ dx = \frac{du}{4x(2x^2 - 1)}. \]

### Step 5: Substitute in the Integral

Now we can substitute \( u \) back into the integral. The first integral becomes:

\[ \int \frac{1}{x \sqrt{u}} \cdot \frac{du}{4x(2x^2 - 1)}. \]

### Step 6: Evaluate the Integral

Now we can evaluate the integral. The first integral simplifies to:

\[ \frac{1}{4} \int \frac{1}{\sqrt{u}} \, du = \frac{1}{4} \cdot 2\sqrt{u} + C = \frac{\sqrt{u}}{2} + C. \]

### Step 7: Back Substitute

Now we substitute back \( u = 2x^4 - 2x^2 + 1 \):

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

### Final Answer

Thus, the evaluated integral is:

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