Question

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

Options

• \[- \frac{1}{\log_e 2}\]

• − log_e 2

• `-1`

• \[\frac{1}{2}\]

Answer 1

[- \frac{1}{\log_e 2}\]

 

\[\text{If }\int\frac{2^\frac{1}{x}}{x^2}dx = k \cdot 2^\frac{1}{x} + C .............(1) \]
\[\text{Let }\frac{1}{x} = t\]
\[ \Rightarrow \frac{- 1}{x^2}dx = dt\]
\[ \Rightarrow \frac{dx}{x^2} = - dt\]
\[\text{Putting }\frac{1}{x} = t\text{ and }\frac{dx}{x^2} = - dt\text{ in LHS of eq. (1), we get}\]
\[ - \int 2^t \cdot dt\]
\[ \Rightarrow - \frac{2^t}{\ln 2} + C\]
\[ \Rightarrow - \frac{2^\frac{1}{x}}{\ln 2} + C . . . (2) \]
\[\text{Comparing RHS of eq (1) with eq (2) we get} , \]
\[ \therefore k = - \frac{1}{\ln 2} or - \frac{1}{\log_e 2}\]