If `int (2^(1//x))/(x^(2))dx= k.2^(1//x)` , then k is equal to :

Options

A. `(-1)/(log 2)`
B. `- log 2`
C. `-1`
D. `(1)/(2)`

Answer

To solve the integral \( \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = k \cdot 2^{\frac{1}{x}} \), we will follow these steps:

### Step 1: Substitution
Let \( t = \frac{1}{x} \). Then, we have:
\[ x = \frac{1}{t} \quad \text{and} \quad dx = -\frac{1}{t^2} \, dt \]

### Step 2: Change the integral
Substituting \( x \) and \( dx \) into the integral:
\[ \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = \int \frac{2^t}{\left(\frac{1}{t}\right)^2} \left(-\frac{1}{t^2}\right) \, dt \]
This simplifies to:
\[ \int 2^t \cdot t^2 \cdot \left(-\frac{1}{t^2}\right) \, dt = -\int 2^t \, dt \]

### Step 3: Integrate
Now we can integrate \( -\int 2^t \, dt \):
\[ -\int 2^t \, dt = -\frac{2^t}{\log 2} + C \]

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

### Step 5: Compare with the given equation
We have:
\[ \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = -\frac{2^{\frac{1}{x}}}{\log 2} + C \]
According to the problem, this equals \( k \cdot 2^{\frac{1}{x}} \). Therefore, we can compare:
\[ k \cdot 2^{\frac{1}{x}} = -\frac{2^{\frac{1}{x}}}{\log 2} \]

### Step 6: Solve for \( k \)
Dividing both sides by \( 2^{\frac{1}{x}} \) (assuming \( 2^{\frac{1}{x}} \neq 0 \)):
\[ k = -\frac{1}{\log 2} \]

### Conclusion
Thus, the value of \( k \) is:
\[ k = -\frac{1}{\log 2} \]