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 start with the integral \( I_n = \int x^n e^{6x} \, dx \) and need to simplify the expression \( 6I_{10} + 10I_{9} \).

### Step 1: Calculate \( I_{10} \)

Using integration by parts, we set:
- \( u = x^{10} \) (algebraic function)
- \( dv = e^{6x} \, dx \) (exponential function)

Then, we differentiate and integrate:
- \( du = 10x^9 \, dx \)
- \( v = \frac{1}{6} e^{6x} \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_{10} = \int x^{10} e^{6x} \, dx = x^{10} \cdot \frac{1}{6} e^{6x} - \int \frac{1}{6} e^{6x} \cdot 10x^9 \, dx \]
\[ = \frac{x^{10} e^{6x}}{6} - \frac{10}{6} I_{9} \]
\[ = \frac{x^{10} e^{6x}}{6} - \frac{5}{3} I_{9} \]

### Step 2: Calculate \( 6I_{10} + 10I_{9} \)

Now we substitute \( I_{10} \) into the expression \( 6I_{10} + 10I_{9} \):
\[ 6I_{10} = 6\left(\frac{x^{10} e^{6x}}{6} - \frac{5}{3} I_{9}\right) \]
\[ = x^{10} e^{6x} - 10 I_{9} \]

Now, substituting this into the expression:
\[ 6I_{10} + 10I_{9} = \left(x^{10} e^{6x} - 10 I_{9}\right) + 10I_{9} \]
\[ = x^{10} e^{6x} \]

### Step 3: Add the constant of integration

Since we are dealing with indefinite integrals, we must include a constant of integration \( c \):
\[ 6I_{10} + 10I_{9} = x^{10} e^{6x} + c \]

### Final Answer

Thus, the expression simplifies to:
\[ \boxed{x^{10} e^{6x} + c} \]