`int(x^6 (x +1))/(sqrt((5x^10) + 6x^9 + x^4)) dx` =

Options

A. `(sqrt((5x^6 + 6x^5 + 1)))/(15)`
B. `(sqrt((5x^8 + 6x^7 + x^2)))/(30)`
C. `(sqrt((5x^8 + 6x^7 + x^2)))/(15)`
D. `sqrt((5x^10 + 6x^5 + x^4)`

Answer

To solve the integral

\[ \int \frac{x^6 (x + 1)}{\sqrt{5x^{10} + 6x^9 + x^4}} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Denominator
First, we notice that we can factor out \(x^4\) from the denominator:

\[ \sqrt{5x^{10} + 6x^9 + x^4} = \sqrt{x^4(5x^6 + 6x^5 + 1)} = x^2 \sqrt{5x^6 + 6x^5 + 1}. \]

### Step 2: Rewrite the Integral
Now, substituting this back into the integral, we have:

\[ \int \frac{x^6 (x + 1)}{x^2 \sqrt{5x^6 + 6x^5 + 1}} \, dx = \int \frac{x^4 (x + 1)}{\sqrt{5x^6 + 6x^5 + 1}} \, dx. \]

### Step 3: Substitute for Simplification
Let us set

\[ t = 5x^6 + 6x^5 + 1. \]

Now, we need to find \(dt\):

\[ dt = (30x^5 + 30x^4) \, dx = 30x^4 (x + 1) \, dx. \]

Thus, we can express \(dx\) as:

\[ dx = \frac{dt}{30x^4 (x + 1)}. \]

### Step 4: Substitute in the Integral
Substituting \(t\) and \(dx\) into the integral, we get:

\[ \int \frac{x^4 (x + 1)}{\sqrt{t}} \cdot \frac{dt}{30x^4 (x + 1)} = \int \frac{1}{30 \sqrt{t}} \, dt. \]

### Step 5: Integrate
Now, we can integrate:

\[ \int \frac{1}{30 \sqrt{t}} \, dt = \frac{1}{30} \cdot 2\sqrt{t} = \frac{2}{30} \sqrt{t} = \frac{1}{15} \sqrt{t} + C. \]

### Step 6: Substitute Back for \(t\)
Finally, we substitute back \(t = 5x^6 + 6x^5 + 1\):

\[ \frac{1}{15} \sqrt{5x^6 + 6x^5 + 1} + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ \int \frac{x^6 (x + 1)}{\sqrt{5x^{10} + 6x^9 + x^4}} \, dx = \frac{1}{15} \sqrt{5x^6 + 6x^5 + 1} + C. \]