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

Options

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

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the Denominator
First, we simplify the expression under the square root in 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, we can rewrite the integral:

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

### Step 3: Use Substitution
Let

\[ t^2 = 5x^6 + 6x^5 + 1. \]

Now, differentiate both sides with respect to \(x\):

\[ 2t \frac{dt}{dx} = 30x^5 + 30x^4 \implies \frac{dt}{dx} = \frac{30x^5 + 30x^4}{2t} = \frac{15(x^5 + x^4)}{t}. \]

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

\[ dx = \frac{2t}{15(x^5 + x^4)} dt. \]

### Step 4: Substitute into the Integral
Substituting \(dx\) back into the integral, we have:

\[ I = \int \frac{x^4 (x + 1) \cdot \frac{2t}{15(x^5 + x^4)} dt}{\sqrt{t^2}}. \]

Since \(\sqrt{t^2} = t\) (assuming \(t > 0\)), we can simplify:

\[ I = \int \frac{2x^4 (x + 1)}{15(x^5 + x^4)} dt. \]

The \(x^4\) terms will cancel out:

\[ I = \int \frac{2(x + 1)}{15} dt. \]

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

\[ I = \frac{2}{15} \int (x + 1) dt. \]

### Step 6: Resubstitute
Finally, we substitute back \(t\):

\[ I = \frac{2}{15} (t + C) = \frac{2}{15} \left(\sqrt{5x^6 + 6x^5 + 1} + C\right). \]

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

\[ I = \frac{2}{15} \sqrt{5x^6 + 6x^5 + 1} + C. \]