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

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

we will follow these steps:

### Step 1: Rewrite the Integral
We start with the given integral:

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

### Step 2: Multiply and Divide by \(x^t\)
To simplify the expression, we multiply and divide by \(x^t\):

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

This gives us:

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

### Step 3: Substitute \(u\)
Let

\[ u = 2x^{3t} + 3x^{2t} + 6x^{t}. \]

Now we differentiate \(u\) with respect to \(x\):

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

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

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

### Step 5: Substitute back into the Integral
Now substitute \(u\) and \(dx\) back into the integral:

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

### Step 6: Simplify the Integral
Now we can simplify the integral further. The integral now becomes:

\[ I = \frac{1}{6t} \int u^{\frac{1}{t}} \, du. \]

### Step 7: Integrate
Now we integrate:

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

### Step 8: Substitute \(u\) back
Substituting back \(u\):

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

### Final Result
Thus, the final result for the integral is:

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