If `I=intx^(27)(6x^(2)+5x+4)(x^(2)+x+1)^(6)dx=f(x)+C`, then f(x) is equal to

Options

A. `(x^(4)(1+x+x^(2))^(7))/(7)+C`
B. `(x^(28)(1+x+x^(2))^(7))/(7)+C`
C. `(x^(28)(1+x+x^(2))^(7))/(28)+C`
D. None

Explanation

To solve the integral \( I = \int x^{27} (6x^2 + 5x + 4)(x^2 + x + 1)^6 \, dx \) and express it in the form \( I = f(x) + C \), we can follow these steps:

### Step 1: Rearranging the Integral
We can rewrite the integral by separating \( x^{27} \) as \( x^{24} \cdot x^3 \):
\[ I = \int x^{24} (6x^2 + 5x + 4)(x^2 + x + 1)^6 \, dx \]

### Step 2: Factor out \( x^{24} \)
Next, we factor \( x^{24} \) out of the polynomial:
\[ I = \int x^{24} (x^3)(6x^2 + 5x + 4)(x^2 + x + 1)^6 \, dx \]

### Step 3: Substitution
We will use the substitution:
\[ t = x^6 + x^5 + x^4 \]
Now, we need to find \( dt \):
\[ dt = (6x^5 + 5x^4 + 4x^3) \, dx \]
This means we can express our integral in terms of \( t \).

### Step 4: Rewrite the Integral
Now we can rewrite the integral using our substitution:
\[ I = \int t^6 \, dt \]

### Step 5: Integrate
Using the power rule for integration:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]
we find:
\[ I = \frac{t^7}{7} + C \]

### Step 6: Substitute Back
Substituting back for \( t \):
\[ I = \frac{(x^6 + x^5 + x^4)^7}{7} + C \]

### Step 7: Simplifying
We can factor out \( x^4 \) from the expression:
\[ I = \frac{x^{28}(1 + x + \frac{1}{x^2})^7}{7} + C \]

### Final Expression for \( f(x) \)
Thus, we can express \( f(x) \) as:
\[ f(x) = \frac{(x^6 + x^5 + x^4)^7}{7} \]