`int (27 e ^(9x) + e ^( 12 x )) ^(1//3) dx` is equal to

Options

A. `(1//4) (27 + e ^(3x)) ^(1//2) +C`
B. `(1//4) (27 + e ^(3x)) ^(2//3) +C`
C. `(1//3) (27 + e ^(3x)) ^(4//3) +C`
D. `(1//4) (27 + e ^(3x)) ^(4//3) +C`

Answer

To solve the integral \( \int (27 e^{9x} + e^{12x})^{1/3} \, dx \), we will follow these steps:

### Step 1: Simplify the expression inside the integral
We start with the integral:
\[ I = \int (27 e^{9x} + e^{12x})^{1/3} \, dx \]
We can factor out \( e^{9x} \) from the expression inside the parentheses:
\[ I = \int \left( e^{9x} (27 + e^{3x}) \right)^{1/3} \, dx \]
Using the property of exponents, we can rewrite this as:
\[ I = \int e^{3x} (27 + e^{3x})^{1/3} \, dx \]

### Step 2: Substitution
Let \( y = 27 + e^{3x} \). Then, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3 e^{3x} \]
This implies:
\[ dx = \frac{dy}{3 e^{3x}} = \frac{dy}{3(y - 27)} \]
Now substituting \( e^{3x} = y - 27 \) into the integral gives:
\[ I = \int (y - 27) \cdot y^{1/3} \cdot \frac{dy}{3(y - 27)} \]
This simplifies to:
\[ I = \frac{1}{3} \int y^{1/3} \, dy \]

### Step 3: Integrate
Now we can integrate \( y^{1/3} \):
\[ I = \frac{1}{3} \cdot \frac{y^{4/3}}{4/3} + C = \frac{1}{4} y^{4/3} + C \]

### Step 4: Substitute back for \( y \)
Now we substitute back \( y = 27 + e^{3x} \):
\[ I = \frac{1}{4} (27 + e^{3x})^{4/3} + C \]

### Final Answer
Thus, the integral is:
\[ \int (27 e^{9x} + e^{12x})^{1/3} \, dx = \frac{1}{4} (27 + e^{3x})^{4/3} + C \]