Answer
$$\frac{{{2^{ - 1/t}}}}{{\ln 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{2^{ - 1/t}}}}{{{t^2}}}} dt \cr
& {\text{Integrate using the substitution method}} \cr
& {\text{Let }}u = - \frac{1}{t},{\text{ }}du = \frac{1}{{{t^2}}}dt,{\text{ }}dt = {t^2}du \cr
& {\text{Substituting}} \cr
& \int {\frac{{{2^{ - 1/t}}}}{{{t^2}}}} dt = \int {\frac{{{2^u}}}{{{t^2}}}} \left( {{t^2}du} \right) \cr
& = \int {{2^u}du} \cr
& {\text{ = }}\frac{{{2^u}}}{{\ln 2}} + C \cr
& {\text{Write in terms of }}t \cr
& {\text{ = }}\frac{{{2^{ - 1/t}}}}{{\ln 2}} + C \cr} $$
