Antiderivatives such as ∫ d x e 2 x + c

Step 1
Firstly, the integral $\int \frac{1}{u^2-c^2}du=-\frac{1}{c}\mathrm{artanh}\frac{u}{c}+C$ is quite standard, so really if you'd have known it your method would be pretty efficient!
Step 2
Secondly, you could first make the substitution $u=e^x/\sqrt{c}$, so that $e^x=u\sqrt{c}$ and $du/dx=e^x/\sqrt{c}=u⟹dx=du/u$. This means that if I is the integral then $\begin{array}{rl}I& =\int \frac{1}{\sqrt{e^{2x}+c}}dx\\ & =\int \frac{1}{\sqrt{u^{2}c+c}}\times \frac{1}{u}du\\ & =\frac{1}{\sqrt{c}}\int \frac{1}{u\sqrt{u^2+1}}du\end{array}$
Here we have a not-very-standard integral: $\int \frac{1}{u\sqrt{u^2+1}}du=-\mathrm{arcosech}u+C=-\mathrm{arsinh}\frac{1}{u}+C$

Step 1
You can write this integral as $I=\int \frac{e^{-x}dx}{\sqrt{1+c{e}^{-2x}}}$
and make the substitution (for $c>0$):
$\sqrt{c}{e}^{-x}=\sinh t$
$\sqrt{1+c{e}^{-2x}}=\cosh t$
$-\sqrt{c}{e}^{-x}dx=\cosh tdt$
Step 2
Then you immediately get
$I=-\frac{1}{\sqrt{c}}\int dt=-\frac{1}{\sqrt{c}}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}(\sqrt{c}{e}^{-x})+const.$.
For $c<0$ there's a similar substitution, $\sqrt{-c}{e}^{-x}=\sin t$