Integrating $\int \frac{e^x}{\sqrt{e^{2x}-1}}dx$ with $u$-substitution only I am trying to integrate this with $u$-substitution only
$$\int \frac{e^x}{\sqrt{e^{2x}-1}}dx$$
I looked at my book's answer key and played around with it a bit, but I'm still very confused.
I know I should choose $u$ to be ${\sqrt{e^{2x}-1}}$
When I differentiate that, I get $$\frac{e^{2x}}{\sqrt{e^{2x}-1}}$$
I get lost here, especially with the numerator. Also is there another $u$ value I can choose?
 A: Using $u=e^x$, $du=e^xdx$ so the integral is equal to
$$\int \frac{du}{\sqrt{u^2-1}}=\cosh^{-1}{(u)}+c$$
$$\therefore \int\frac{e^x}{\sqrt{e^{2x}-1}}dx=\cosh^{-1}{(e^x)}+c$$
A: With $u = \sqrt{e^{2x} - 1}$, you have $e^x = \sqrt{u^2 + 1}$. Then 
$$du = \frac{e^{2x}}{\sqrt{e^{2x} - 1}} dx = \frac{u^2 + 1}{u}du \implies dx = \frac{u}{u^2 + 1} du$$
Then
$$\int \frac{e^x}{\sqrt{e^{2x} - 1}} dx = \int \frac{\sqrt{u^2 + 1}}{u} \cdot \frac{u}{u^2 + 1} du = \int \frac{1}{\sqrt{u^2 + 1}}du$$
so I'm not exactly seeing where the issue lies with that substitution; maybe a miscalculation on your part? This seems to match the form of integral $(32)$ on this reference sheet I have bookmarked, on the off chance it's actually a form you're unfamiliar with (though the answer is more succinctly sometimes seen in terms of hyperbolic arccosine). 
The result would be, regardless,
$$\int \frac{1}{\sqrt{u^2 + 1}}du = \ln \left| u + \sqrt{u^2 + 1} \right| + C = \cosh^{-1}(\sqrt{u^2 + 1}) + C$$
and then you just bring back in your $u$. This gives $\sqrt{e^{2x} - 1} + e^x$ inside the logarithm and thus
$$\int \frac{e^x}{\sqrt{e^{2x} - 1}} dx = \ln \left| e^x + \sqrt{e^{2x} + 1} \right| + C = \cosh^{-1}(e^x) +C$$

Footnote: Peter Foreman's answer reaches the same conclusion, and is just as valid. You could use either $u$-substitution, whichever you think is easier to contend with.
A: $$\text{Let }\begin{bmatrix}u \\ \mathrm du\end{bmatrix}=\begin{bmatrix}e^x\\e^x\mathrm dx\end{bmatrix}\implies \int\dfrac{e^x}{\sqrt{e^{2x}-1}}\mathrm dx=\int\dfrac{1}{\sqrt{u^2-1}}\mathrm du$$
As has been already illustrated by @EeveeTrainer and @PeterForeman the techniques of u-substitution and knowing-the-derivative-of $\cosh^{-1}x$. So, I would present another way to tackle this using trig-substitution. 
$$\text{Let }\begin{bmatrix}u\\\mathrm du\end{bmatrix}=\begin{bmatrix}\sec\theta\\\sec\theta\tan\theta\mathrm d\theta\end{bmatrix}\implies \int \dfrac{1}{\sqrt{u^2-1}}\mathrm du=\int \dfrac{\sec\theta\tan\theta}{\sec\theta}\mathrm d\theta$$
Can you proceed?
