∫ e 4 x + e 2 x d x =

We have $I=\int \sqrt{e^{4x}+e^{2x}}dx=\int \sqrt{e^{4x}+e^{2x}}dx$$=\int e^x\sqrt{{\left(e^x\right)}^2+1}dx$

Substitute, $e^x=v \Rightarrow dv=e^{x}dx$

$I=\int \sqrt{v^2+1}dv$

Now assume $v=\tan t\Rightarrow dv={\sec }^{2}t·dt$

$I=\int \sqrt{{\tan }^{2}t+1}·{\sec }^{2}t·dt=\int {\sec }^3 t·dt$

$=\int {\sec }^3 t·dt=\int \sec t·{\sec }^{2}t·dt=\sec t\int {\sec }^{2}tdt-\int \left(\frac{d\left(\mathrm{se}c t\right)}{dt}\int {\sec }^{2}tdt\right)dt$

$=\sec t·\tan t-\int \left(\sec t·{\tan }^{2}t\right)dt=\sec t·\tan t-\int \left(\sec t·\left({\sec }^{2}t-1\right)\right)dt$$=\sec t·\tan t-\int {\sec }^{3}t·dt+\int \sec t·dt=\sec t·\tan t-I+\int \sec t·dt$

$\Rightarrow 2I=\sec t·\tan t+\int \sec t·dt=\sec t·\tan t+\ln \left|\sec t+\tan t\right|+c$

$\Rightarrow I=\frac{1}{2}\sec t·\tan t+\frac{1}{2}\ln \left|\sec t+\tan t\right|+c$

Now substitute $\tan t=e^x$

we get, 

$\Rightarrow I=\frac{1}{2}{e}^x\left(\sqrt{e^{2x}+1}\right)+\frac{1}{2}\ln \left|e^x+\sqrt{e^{2x}+1}\right|+c$

$\Rightarrow I=\frac{1}{2}{e}^x\left(\sqrt{e^{2x}+1}\right)+\frac{1}{2}\sin h^{-1}\left(e^x\right)+c$