For any natural number `t,int(x^(3t)+x^(2t)+x^(t))(2x^(2t)+3x^(t)+6)^(1//t)dx` is :

Options

A. `(1)/(6(t+1))(2x^(3t)+3x^(2t)+6x^(t))^((t+1)/(t))+C`
B. `(1)/(6t)(2x^(2t)+3x^(t)+6x)^(t)+C`
C. `(1)/(t+1)(x^(3t)+x^(2t)+x^(t))^(2)+C`
D. None of these

Answer

To solve the integral

\[ I = \int \left( x^{3t} + x^{2t} + x^{t} \right) \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t}} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral in a more manageable form. Let's denote:

\[ u = \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t}}. \]

Thus, we can express the integral as:

\[ I = \int \left( x^{3t} + x^{2t} + x^{t} \right) u \, dx. \]

### Step 2: Simplify the Expression
Next, we can multiply the first part of the integrand by \(x^{t}\) and divide by \(x^{t}\):

\[ I = \int \frac{x^{3t} + x^{2t} + x^{t}}{x^{t}} \cdot u \cdot x^{t} \, dx = \int \left( x^{2t} + x^{t} + 1 \right) u \cdot x^{t} \, dx. \]

### Step 3: Substitute \(u\) and Differentiate
Now, we differentiate \(u\) with respect to \(x\):

\[ u = \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t}}. \]

Using the chain rule, we find:

\[ \frac{du}{dx} = \frac{1}{t} \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t} - 1} \cdot \left( 4tx^{2t-1} + 3tx^{t-1} \right). \]

### Step 4: Express \(dx\) in terms of \(du\)
From the above, we can express \(dx\) as:

\[ dx = \frac{t}{\left( 4tx^{2t-1} + 3tx^{t-1} \right) \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t} - 1}} \, du. \]

### Step 5: Substitute back into the Integral
Substituting \(dx\) back into the integral gives:

\[ I = \int \left( x^{2t} + x^{t} + 1 \right) u \cdot \frac{t}{\left( 4tx^{2t-1} + 3tx^{t-1} \right) \left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t} - 1}} \, du. \]

### Step 6: Solve the Integral
Now, we can integrate with respect to \(u\):

\[ I = \frac{t}{4t} \int u^{\frac{1}{t}} \, du = \frac{t}{4t} \cdot \frac{u^{\frac{1}{t} + 1}}{\frac{1}{t} + 1} + C. \]

### Step 7: Back Substitute \(u\)
Finally, substitute back \(u\):

\[ I = \frac{t}{4t} \cdot \frac{\left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t} + 1}}{\frac{1}{t} + 1} + C. \]

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

\[ I = \frac{t}{4t} \cdot \frac{\left( 2x^{2t} + 3x^{t} + 6 \right)^{\frac{1}{t} + 1}}{\frac{1}{t} + 1} + C. \]