Odpowiedź:
[tex]\huge\boxed{\int\dfrac{2^x}{\sqrt{1-4^x}}dx=\dfrac{\arcsin2^x}{\ln2}+C}[/tex]
Szczegółowe wyjaśnienie:
[tex]\int\dfrac{2^x}{\sqrt{1-4^x}}dx\Rightarrow\left[\begin{array}{ccc}2^x=t\\2^x\ln2\ dx=dt\\2^x\ dx=\dfrac{1}{\ln2}dt\end{array}\right] \Rightarrow\int\dfrac{1}{\sqrt{1-t^2}}\cdot\dfrac{1}{\ln2}dt=\dfrac{1}{\ln2}\arcsin t\\\\=\dfrac{\arcsin2^x}{\ln2}[/tex]