If `int(2^(1//x))/(x^(2))dx=a2^(1//x)+C`, then a=

Options

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

Explanation

To solve the integral \( \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = a \cdot 2^{\frac{1}{x}} + C \), 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: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral gives:
\[ \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = \int \frac{2^t}{\left(\frac{1}{t}\right)^2} \left(-\frac{1}{t^2}\right) dt \]
This simplifies to:
\[ \int 2^t \cdot t^2 \cdot \left(-\frac{1}{t^2}\right) dt = -\int 2^t \, dt \]

### Step 3: Integrate
The integral of \( 2^t \) is:
\[ -\int 2^t \, dt = -\frac{2^t}{\ln 2} + C \]

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

### Step 5: Compare with Given Expression
We have:
\[ -\frac{2^{\frac{1}{x}}}{\ln 2} + C = a \cdot 2^{\frac{1}{x}} + C \]
By comparing coefficients, we find:
\[ a = -\frac{1}{\ln 2} \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{-\frac{1}{\ln 2}} \]