If ` int (2^(x))/(sqrt(1-4^(x))) dx = k sin ^(-1) (f(x)) + C` then :

Options

A. ` k = log 2, f(x) = 2^(x)`
B. `k = (1)/(log 2) , f(x) = 2^(x)`
C. `k = log 2, f(x) = 4^(x)`
D. `k = (1)/(log2) , f(x) = 4^(x)`

Answer

To solve the integral \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \) and express it in the form \( k \sin^{-1}(f(x)) + C \), we will follow these steps:

### Step 1: Substitution
Let \( t = 2^x \). Then, we have:
\[ 4^x = (2^2)^x = (2^x)^2 = t^2 \]
Also, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{2^x \log 2} = \frac{dt}{t \log 2} \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \int \frac{t}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \log 2} \]
This simplifies to:
\[ \frac{1}{\log 2} \int \frac{1}{\sqrt{1 - t^2}} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{\sqrt{1 - t^2}} \, dt \) is a standard integral that results in:
\[ \sin^{-1}(t) + C \]
Thus, we have:
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \frac{1}{\log 2} \left( \sin^{-1}(t) + C \right) \]

### Step 4: Substitute Back
Now, substituting back \( t = 2^x \):
\[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \frac{1}{\log 2} \sin^{-1}(2^x) + C \]

### Step 5: Identify \( k \) and \( f(x) \)
From the expression \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = k \sin^{-1}(f(x)) + C \), we can identify:
- \( k = \frac{1}{\log 2} \)
- \( f(x) = 2^x \)

### Final Answer
Thus, the values are:
- \( k = \frac{1}{\log 2} \)
- \( f(x) = 2^x \)