Question

somebody can help me to integrate this?
2^x /(4+4^x) dx

Answer 1

Have you noticed that \(4^{x} = 2^{2x} = \left(2^{x}\right)^{2}\)?

Answer 2

lets do some substitutions
\[u = 2^x\]
\[du = \ln(2) 2^x dx\]
the 2^x on top will cancel leaving
\[\frac{1}{\ln 2} \int\limits \frac{du}{4 + u^2}\]
now use identity --> 1 +tan^2 = sec^2, to help make next substitution
\[u = 2 \tan \theta\]
\[du = 2 \sec^2 \theta\]
\[\rightarrow \frac{1}{\ln 2} \int\limits \frac{2 \sec^2 \theta}{4 + 4\tan^2 \theta}\]
from identity, the sec^2 will cancel leaving
\[\frac{1}{2 \ln 2} \int\limits d \theta\]
\[= \frac{1}{2 \ln 2} \theta = \frac{1}{2 \ln 2} \tan^{-1} \frac{u}{2}\]
\[=\frac{1}{2 \ln 2} \tan^{-1} (\frac{2^x}{2})\]