$\int\frac1{e^x\sqrt{4+e^{2x}}}dx$
In my textbook the solution of this integral says that $u=e^x$ and $du=e^xdx$ is the substitution to go, and then transforms the integer in this
$\int\frac1{u\sqrt{4+u^{2x}}}du$
but according to my knowledge, it should be like this
$\int\frac1{u^{2}\sqrt{4+u^{2x}}}du$
because $\frac{du}{u}=dx$, so you have $\int\frac1{u\sqrt{4+u^{2x}}}*\frac{du}{u}$
did i understand wrong the substitution?
Should be $\int{du\over{u^2\sqrt{4+u^2}}}$ with that substitution. Replace $dx$ with $du\over{e^x}$ and $e^{2x}$ with $u^2$
The x is substituted out when $e^{2x}$ is substituted by $u^2$.