If $\int \frac{2^{1 / x}}{x^2} d x=a 2^{1 / x}+C$, then $a=$
Answer & explanation
Correct answer: option 1
We have,
$\int \frac{2^{1 / x}}{x^2} d x=-\int 2^{1 / x} d\left(\frac{1}{x}\right)=\frac{-2^{1 / x}}{\log _e 2}+C=-\left(\log _2 e\right) 2^{1 / x}+C$
∴ $a=-\log _2 e$