`int(x^(7m)+x^(2m)+x^m)(2x^(6m)+7x^m+14)^(1/m)dx`

Options

A. `(7x^(7m)+2x^(2m)+14x^(m))^((m+1)/(m))/(14(m+1))+C`
B. `(2x^(7m)+14x^(2m)+7x^(m))^((m+1)/(m))/(14(m+1))+C`
C. `(2x^(7m)+7x^(2m)+14x^(m))^((m+1)/(m))/(14(m+1))+C`
D. `(7x^(7m)+2x^(2m)+x^(m))^((m+1)/(m))/(14(m+1))+C`

Explanation

To solve the integral

\[ \int (x^{7m} + x^{2m} + x^m)(2x^{6m} + 7x^m + 14)^{\frac{1}{m}} \, dx, \]

we will follow these steps:

### Step 1: Factor out \( x \)

First, we can factor out \( x \) from the first part of the integrand:

\[ x^{7m} + x^{2m} + x^m = x^{m}(x^{6m} + x^{m} + 1). \]

Thus, we rewrite the integral as:

\[ \int x^{m}(x^{6m} + x^{m} + 1)(2x^{6m} + 7x^m + 14)^{\frac{1}{m}} \, dx. \]

### Step 2: Substitute \( t \)

Next, we make a substitution for the second part of the integrand. Let:

\[ t = 2x^{7m} + 7x^{2m} + 14x^m. \]

Now, we differentiate \( t \):

\[ \frac{dt}{dx} = 14mx^{7m-1} + 14mx^{m-1}. \]

This implies:

\[ dt = (14mx^{7m-1} + 14mx^{m-1}) \, dx. \]

### Step 3: Express \( dx \)

Rearranging gives:

\[ dx = \frac{dt}{14m(x^{7m-1} + x^{m-1})}. \]

### Step 4: Substitute back into the integral

Now substituting back into the integral, we have:

\[ \int x^{m}(x^{6m} + x^{m} + 1)(2x^{7m} + 7x^{2m} + 14x^m)^{\frac{1}{m}} \cdot \frac{dt}{14m(x^{7m-1} + x^{m-1})}. \]

### Step 5: Simplify the integral

The integral simplifies to:

\[ \frac{1}{14m} \int t^{\frac{1}{m}} \, dt. \]

### Step 6: Integrate

Using the formula for integration:

\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C, \]

we find:

\[ \frac{1}{14m} \cdot \frac{t^{\frac{1}{m} + 1}}{\frac{1}{m} + 1} + C. \]

### Step 7: Substitute back \( t \)

Substituting back \( t = 2x^{7m} + 7x^{2m} + 14x^m \):

\[ = \frac{1}{14m} \cdot \frac{(2x^{7m} + 7x^{2m} + 14x^m)^{\frac{1}{m} + 1}}{\frac{1}{m} + 1} + C. \]

### Step 8: Final Simplification

This gives us:

\[ = \frac{(2x^{7m} + 7x^{2m} + 14x^m)^{\frac{1}{m} + 1}}{14(m + 1)} + C. \]

### Final Answer

Thus, the final answer is:

\[ \frac{(2x^{7m} + 7x^{2m} + 14x^m)^{\frac{1}{m} + 1}}{14(m + 1)} + C. \]