Let \(u = \frac{1}{x} = x^{-1}\)
Differentiating \(u\) with respect to \(x\) :
\( du = -\frac{1}{x^2} dx\)
Substitute \(u\) and \(du\) into the integral:
\(\displaystyle \int \frac{1}{x^2} 2^{\frac{1}{x}} dx = \int 2^u (-du) = - \int 2^u du\)
Using the standard integral formula \(\displaystyle \int a^u du = \frac{a^u}{\ln a} + C\) (with \(a=2\)):
\(\displaystyle - \int 2^u du = - \left( \frac{2^u}{\ln 2} \right) + C\)
Hence
\(\displaystyle \int \frac{1}{x^2} 2^{\frac{1}{x}} dx = - \frac{1}{\ln 2} \cdot 2^{\frac{1}{x}} + C\)
Determining the Value of \(k\)
By comparing our result with the given form \(\displaystyle k \cdot 2^{\frac{1}{x}} + C\):
\(\displaystyle k \cdot 2^{\frac{1}{x}} + C = - \frac{1}{\ln 2} \cdot 2^{\frac{1}{x}} + C\)
The value of \(k\) is:
\(\displaystyle k = - \frac{1}{\ln 2}\)
Since \(\log 2\) in the options typically denotes the natural logarithm \(\ln 2\) in calculus, the answer is:
\(\displaystyle \frac{-1}{\log 2}\)