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