Question

If $\int \frac{2^{1/x}}{x^2}dx=k{2}^{1/x}+C,$ then k is equal to

(a) $-\frac{1}{{\log }_{e}2}$

(b) − log_e 2

(c) − 1

(d) $\frac{1}{2}$

Answer 1

(a) $-\frac{1}{{\log }_{e}2}$

$$\begin{gathered} \mathrm{If}\int \frac{2^{{\displaystyle \frac{1}{x}}}}{x^2}dx=k·2^{\frac{1}{x}}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}...\left(1\right) \\ \mathrm{Let}\frac{1}{x}=t \\ \Rightarrow \frac{-1}{x^2}dx=dt \\ \Rightarrow \frac{dx}{x^2}=-dt \\ \mathrm{Putting}\frac{1}{x}=t\mathrm{and}\frac{dx}{x^2}=-dt\mathrm{in}\mathrm{LHS}\mathrm{of}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ -\int 2^t·dt \\ \Rightarrow -\frac{2^t}{\ln 2}+C \\ \Rightarrow -\frac{2^{{\displaystyle \frac{1}{x}}}}{\ln 2}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{} \\ \mathrm{Comparing}\mathrm{RHS}\mathrm{of}\mathrm{eq}\left(1\right)\mathrm{with}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}, \\ \therefore k=-\frac{1}{\ln 2}\mathrm{or}-\frac{1}{{\log }_{e}2} \end{gathered}$$