If $\int\frac{2^x}{\sqrt{1-4^x}}dx=k\sin^{-1}(f(x))+C$ then:

$k=\log 2,f(x)=2^x$ $k=\frac{1}{\log 2},f(x)=2^x$ $k=\log 2,f(x)=4^x$ $k=\frac{1}{\log 2},f(x)=4^x$

$k=\frac{1}{\log 2},f(x)=2^x$

Let $\int\frac{2^x}{\sqrt{1-4^x}}dx=\frac{1}{\log 2}\int\frac{1}{\sqrt{1-t^2}}dt$ Putting $2^x=t,2^x\log 2\,dx=dt$ $I=\frac{1}{\log 2}\sin^{-1}(\frac{t}{1})+C=\frac{1}{\log 2}\sin^{-1}(2^x)+C⇒k=\frac{1}{\log 2},f(x)=2^x$