Question
$\int \sqrt{4^x\left(4^x+4\right)} d x$

Answer

$ \text { Let } I =\int \sqrt{4^x\left(4^x+4\right)} d x$
$=\int \sqrt{\left(2^x\right)^2\left[\left(2^x\right)^2+4\right]} d x$
$=\int \sqrt{\left(2^x\right)^2+2^2 \cdot 2^x d x} $
Put $2^x=t$
$ \therefore 2^{ x } \log 2 dx = dt$
$\therefore 2^{ x } dx =\frac{1}{\log 2} dt$
$\therefore I =\frac{1}{\log 2} \int \sqrt{ t ^2+2^2} dt$
$=\frac{1}{\log 2}\left[\frac{ t }{2} \sqrt{ t ^2+2^2}+\frac{2^2}{2} \log \left| t +\sqrt{ t ^2+2^2}\right|\right]+ c$
$\therefore I =\frac{1}{\log 2}\left[\frac{2^x}{2} \sqrt{4 x+4}+2 \log \left|2^x+\sqrt{4^x+4}\right|\right]+ c $

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