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 \(\int \frac{2^{\frac{1}{x}}}{x^2} \, dx\) and find the value of \(k\) such that \(\int \frac{2^{\frac{1}{x}}}{x^2} \, dx = k \cdot 2^{\frac{1}{x}}\), we will use substitution. Here are the steps:

### Step 1: Substitution
Let \(y = \frac{1}{x}\). Then, we differentiate both sides:
\[ \frac{dy}{dx} = -\frac{1}{x^2} \Rightarrow dy = -\frac{1}{x^2} \, dx \Rightarrow dx = -x^2 \, dy \]
Since \(x = \frac{1}{y}\), we have:
\[ dx = -\left(\frac{1}{y^2}\right) dy \]

### Step 2: Rewrite the Integral
Substituting \(y\) into the integral, we get:
\[ \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = \int 2^y \left(-\frac{1}{y^2}\right) dy = -\int \frac{2^y}{y^2} \, dy \]

### Step 3: Integrate
Now we need to integrate \(-\int \frac{2^y}{y^2} \, dy\). The integral of \(a^x\) is given by:
\[ \int a^x \, dx = \frac{a^x}{\ln a} + C \]
Thus, we can apply this to our integral:
\[ -\int 2^y \, dy = -\frac{2^y}{\ln 2} + C \]

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

### Step 5: Express in Terms of k
We can express the result as:
\[ \int \frac{2^{\frac{1}{x}}}{x^2} \, dx = -\frac{1}{\ln 2} \cdot 2^{\frac{1}{x}} + C \]
Comparing this with the given form \(k \cdot 2^{\frac{1}{x}}\), we find:
\[ k = -\frac{1}{\ln 2} \]

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