MCQ
If $\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{dx}=\text{k }2^{\frac{1}{\text{x}}}+\text{C},$ then $k$ is equal to:
- ✓$-\frac{1}{\log_\text{e}2}$
- B$-\log_\text{e}2$
- C$-1$
- D$\frac{1}{2}$
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