`int2^((3x+4))dx=?`

Options

A. `(3)/((log2))*2^((3x+4))+C`
B. `(2^((2x+4)))/(3(log2)+C)`
C. `(2)/((log2))*2^((3x+4))+C`
D. none of these

Explanation

To solve the integral \( \int 2^{(3x+4)} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = 3x + 4 \).

### Step 2: Differentiate
Now, differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 3 \implies dx = \frac{dt}{3} \]

### Step 3: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral gives:
\[ \int 2^{(3x+4)} \, dx = \int 2^t \cdot \frac{dt}{3} \]

### Step 4: Factor Out the Constant
We can factor out \( \frac{1}{3} \):
\[ = \frac{1}{3} \int 2^t \, dt \]

### Step 5: Integrate
The integral of \( 2^t \) is:
\[ \int 2^t \, dt = \frac{2^t}{\log 2} + C \]
Thus, we have:
\[ = \frac{1}{3} \left( \frac{2^t}{\log 2} + C \right) \]

### Step 6: Substitute Back
Now substitute back \( t = 3x + 4 \):
\[ = \frac{1}{3} \cdot \frac{2^{(3x + 4)}}{\log 2} + C \]

### Step 7: Final Simplification
This can be simplified to:
\[ = \frac{2^{(3x + 4)}}{3 \log 2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int 2^{(3x+4)} \, dx = \frac{2^{(3x + 4)}}{3 \log 2} + C \]

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