How to integrate it? I've got a problem with this integral.
How to calculate it?
$$
\int\frac{2^x}{\sqrt{1-4^x}} \, dx
$$
Regards.

Here's a hint:
$$
\int\frac{1}{\sqrt{1-4^x}}\Big(2^x\,dx\Big)
$$
In other words, you want a substitution in which $2^x\,dx$ becomes either $du$ or some constant times $du$.  That's the sort of thing you want to look for.
That's the hint.  Below are the details.
\begin{align}
u & = 2^x \\[10pt]
u^2 & = 4^x \\[10pt]
du & = 2^x(\log_e 2)\,dx
\end{align}
So
$$
\int\frac{2^x}{\sqrt{1-4^x}}\, dx = \frac{1}{\log_e2} \int \frac{du}{\sqrt{1-u^2}} =\frac{1}{\log_e 2}\arcsin(u)+C
$$
$$
=\frac{1}{\log_e 2}\arcsin(2^x)+C.
$$
