`int sqrt(4^(x) (4^(x) + 4))dx`

Answer

To solve the integral \( I = \int \sqrt{4^x (4^x + 4)} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ I = \int \sqrt{4^x (4^x + 4)} \, dx \]
We can rewrite \( 4^x \) as \( (2^2)^x = 2^{2x} \). Therefore, we have:
\[ I = \int \sqrt{2^{2x} (2^{2x} + 4)} \, dx \]
Next, we can factor out \( 4 \) from the second term inside the square root:
\[ I = \int \sqrt{2^{2x} (2^{2x} + 2^2)} \, dx \]

### Step 2: Factor out \( 2^{2x} \)
Now we can factor out \( 2^{2x} \) from the square root:
\[ I = \int \sqrt{2^{2x}} \sqrt{2^{2x} + 4} \, dx = \int 2^x \sqrt{2^{2x} + 4} \, dx \]

### Step 3: Substitute \( 2^x \) with \( t \)
Let \( 2^x = t \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{t \ln 2} \]
Substituting this into the integral gives:
\[ I = \int t \sqrt{t^2 + 4} \cdot \frac{dt}{t \ln 2} = \frac{1}{\ln 2} \int \sqrt{t^2 + 4} \, dt \]

### Step 4: Integrate \( \sqrt{t^2 + 4} \)
The integral \( \int \sqrt{t^2 + 4} \, dt \) can be solved using the standard formula:
\[ \int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln \left( x + \sqrt{x^2 + a^2} \right) + C \]
In our case, \( a = 2 \):
\[ \int \sqrt{t^2 + 4} \, dt = \frac{t}{2} \sqrt{t^2 + 4} + 2 \ln \left( t + \sqrt{t^2 + 4} \right) + C \]

### Step 5: Substitute back to \( x \)
Now substituting back \( t = 2^x \):
\[ I = \frac{1}{\ln 2} \left( \frac{2^x}{2} \sqrt{(2^x)^2 + 4} + 2 \ln \left( 2^x + \sqrt{(2^x)^2 + 4} \right) \right) + C \]
This simplifies to:
\[ I = \frac{2^x}{2 \ln 2} \sqrt{4^x + 4} + \frac{2}{\ln 2} \ln \left( 2^x + \sqrt{4^x + 4} \right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{2^x}{2 \ln 2} \sqrt{4^x + 4} + \frac{2}{\ln 2} \ln \left( 2^x + \sqrt{4^x + 4} \right) + C \]