If `I_(n)=intx^(n)e^(6x)dx`, then the expression `6I_(10)+10I_(9)` simplifies to (where, c is the constant of integration)

Options

A. `x^(10)e^(5x)+c`
B. `x^(10)e^(6x)+c`
C. `x^(9)e^(5x)+c`
D. `x^(10)e^(10x)+c`

Answer

To solve the problem, we need to evaluate the expression \( 6I_{10} + 10I_{9} \) where \( I_n = \int x^n e^{6x} \, dx \).

### Step 1: Write down the integrals
We have:
\[ I_{10} = \int x^{10} e^{6x} \, dx \]
\[ I_{9} = \int x^{9} e^{6x} \, dx \]

### Step 2: Use integration by parts for \( I_{10} \)
Using integration by parts, we let:
- \( u = x^{10} \) (algebraic part)
- \( dv = e^{6x} \, dx \) (exponential part)

Then, we differentiate and integrate:
- \( du = 10x^{9} \, dx \)
- \( v = \frac{e^{6x}}{6} \)

Applying integration by parts:
\[ I_{10} = uv - \int v \, du \]
Substituting the values:
\[ I_{10} = x^{10} \cdot \frac{e^{6x}}{6} - \int \frac{e^{6x}}{6} \cdot 10x^{9} \, dx \]
\[ I_{10} = \frac{x^{10} e^{6x}}{6} - \frac{10}{6} I_{9} \]

### Step 3: Multiply the equation by 6
Now, we multiply the entire equation by 6:
\[ 6I_{10} = x^{10} e^{6x} - 10 I_{9} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 6I_{10} + 10I_{9} = x^{10} e^{6x} \]

### Conclusion
Thus, we find that:
\[ 6I_{10} + 10I_{9} = x^{10} e^{6x} \]

### Final Answer
The expression \( 6I_{10} + 10I_{9} \) simplifies to:
\[ \boxed{x^{10} e^{6x}} \]