Help evaluating $\int e^x \sqrt{1+e^{2x}}dx$ $\int e^x \sqrt{1+e^{2x}}dx$
It's probably been answered somewhere, but I havent found it so far so I decided to post it as a question (if it has been answered point me in the right direction and I will delete the question.
What I have tried:


*

*Put it into mathematica, got a relatively simple answer:
$$\frac{1}{2} (e^x \sqrt{1+e^{2x}}+\sinh^{-1}e^x)$$

*Dont know how to get there! Obviously I tried $e^x = u$  substitution, ended up with $\int \sqrt{1+u^2}du$. From here I am not really sure what do to, I tried another substitution of $u=\cosh t$, but to be honest I am very unfamiliar with hyperbolic trig:


$$u=\cosh t \implies du=\sinh t~dt$$
$$\int \sinh t\sqrt{1+\cosh^2t} ~ ~dt$$
And i believe this is the same as $\int \sinh^2 t ~ ~dt$?
Once again, I am not too sure about this because I don't know much about hyperbolic trig. 
From here I don't know where to go (don't even know if my working so far is right)..
 A: $e^x=\tan\theta \implies e^x\ dx=\sec^2\theta\ d\theta$
$\therefore\displaystyle\int e^x\sqrt{e^{2x}+1}\ dx=\displaystyle\int \sec^3\theta\ d\theta=I$
$\begin{align}I & =\displaystyle\int \sec \theta\cdot \sec^2\theta\ d\theta\\ & =\sec\theta\displaystyle\int \sec^2\theta\ d\theta-\displaystyle\int\sec\theta\tan^2\theta\\&=\sec\theta\tan\theta-\color{blue}{\displaystyle\int\sec^3\theta\ d\theta}+\displaystyle\int\sec\theta\ d\theta\\&= \sec\theta\tan\theta-\color{blue}{I}+\displaystyle\int\sec\theta\ d\theta\end{align}$
A: The first step which you got is that $$\int\; f(x)\; dx = \int\; \sqrt{1+u^2} du \\
= \int\; \cosh^2(t) dt = \int\; \frac{\cosh(2t)+1}{2} dt \\
= \frac{\sinh(2t)}{4} + t/2 + c \\
= \frac{\sinh(2\sinh^{-1}(u))}{4} + \frac{\sinh^{-1}(u)}{2} \\
= \frac{i \sin(2i\sin^{-1}(u/i)}{4} + \frac{\sinh^{-1}(e^x)}{2} \\
= \frac{2i\cdot \sin(i\sin^{-1}(u/i))\cdot \cos(i\sin^{-1}(u/i))}{4} + (*) \\
= \frac{2\cdot u\cdot \cosh(\sinh^{-1}(u))}{4} + (*) \\
= \frac{u\cdot \sqrt{u^2+1}}{2} + \frac{\sinh^{-1}(e^x)}{2} \\
= \frac{e^x\cdot \sqrt{e^{2x}+1}}{2} + \frac{\sinh^{-1}(e^x)}{2}$$, where we use that
$\cosh(2t)+1 = 2\cosh^2(t)$ and $\sinh(iz) = i\sin(z)$. Perhaps someone out there memorizes those hyperbolic formulae, but I find it easier to go back to what I know - the trusty sines and cosines. This yields the desired result.
A: You're close.
$u=e^x$, $\mathrm{d}u = e^x\,\mathrm{d}x$
$\int\sqrt{1 + u^2}\,\mathrm{d}u$
$u = \sinh t$, $\mathrm{d}u = \cosh t\,\mathrm{d}t$
$\int\cosh t\sqrt{1 + \sinh^2 t}\,\mathrm{d}t$
$\cosh^2 t = 1 + \sinh^2 t$
$\int\cosh^2 t\,\mathrm{d}t$ = $\int\frac{1}{2}\left(1 + \cosh 2t\right)\mathrm{d}t$
$= \frac{t}{2} + \frac{\sinh 2t}{4}$ = $\frac{1}{2}\left(\mathrm{arcsinh}\,u + u\sqrt{1 + u^2}\right)$ = $\frac{1}{2}\left(e^x\sqrt{1 + e^{2x}} + \mathrm{arcsinh}\,e^x\right)$ whence the result you get from Mathematica.
We use the identity $\sinh\left(2\,\mathrm{arcsinh}\,u\right) = 2u\sqrt{1 + u^2}$
