The given integral is
$$ I=\int 2^{-x}dx \tag1$$

We solve the above integral by using the following substitution:
\begin{align*}
&2^{-x} = t\\
\implies -2^{-x}\cdot&\ln(2)\cdot dx=dt\\
\implies 2^{-x}&\cdot dx =-\frac{dt}{\ln(2)}
\end{align*}

Substituting the above value in the integral (1), we have:
\begin{align*}
I&=\int -\frac{dt}{\ln(2)}\\
&= -\frac{1}{\ln(2)}\int dt\\
&=-\frac{1}{\ln(2)}t +c\\
&= \frac{-2^{-x}}{\ln(2)} +c
\end{align*}
Thus, we have
$$\int 2^{-x}dx = -\frac{2^{-x}}{\ln(2)} +c$$