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
    $\frac{1}{2}$

Answer

Correct option: A.
$-\frac{1}{\log_\text{e}2}$
$\text{k}=\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{ dx}$
Put $\frac{1}{\text{x}}=\text{t}$
$\frac{-1}{\text{x}^2}\text{ dx}=\text{dt}$
$\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
$\text{k}=\int2^{\text{t}}(-\text{dt})$
$\text{k}=\frac{-2^{\text{t}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-2^{\frac{1}{\text{x}}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-1}{\log_\text{e}2}$

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