Evaluate the following integrals
`int (2^(x))/(sqrt(4^(x)-25))dx`

Answer

To evaluate the integral

\[ I = \int \frac{2^x}{\sqrt{4^x - 25}} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the integral
We can express \(4^x\) in terms of \(2^x\):

\[ 4^x = (2^2)^x = (2^x)^2. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{2^x}{\sqrt{(2^x)^2 - 25}} \, dx. \]

### Step 2: Substitution
Let \(t = 2^x\). Then, we differentiate \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = 2^x \ln(2) \implies dx = \frac{dt}{2^x \ln(2)} = \frac{dt}{t \ln(2)}. \]

### Step 3: Substitute in the integral
Now substitute \(t\) and \(dx\) into the integral:

\[ I = \int \frac{t}{\sqrt{t^2 - 25}} \cdot \frac{dt}{t \ln(2)} = \frac{1}{\ln(2)} \int \frac{1}{\sqrt{t^2 - 25}} \, dt. \]

### Step 4: Evaluate the integral
The integral \(\int \frac{1}{\sqrt{t^2 - 25}} \, dt\) can be evaluated using the formula:

\[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C. \]

Here, \(a = 5\). Thus, we have:

\[ \int \frac{1}{\sqrt{t^2 - 25}} \, dt = \ln |t + \sqrt{t^2 - 25}| + C. \]

### Step 5: Substitute back
Substituting back into our expression for \(I\):

\[ I = \frac{1}{\ln(2)} \left( \ln |t + \sqrt{t^2 - 25}| + C \right). \]

Now substituting \(t = 2^x\):

\[ I = \frac{1}{\ln(2)} \left( \ln |2^x + \sqrt{(2^x)^2 - 25}| + C \right). \]

### Final Answer
Thus, the final answer is:

\[ I = \frac{1}{\ln(2)} \ln \left( 2^x + \sqrt{4^x - 25} \right) + C. \]

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