Is there a good strategy for calculating antiderivatives such as$$\int \frac{1}{\sqrt{e^{2x} +c}}dx\enspace?$$
Right now I'm substituting the entire radical expression, which sort of simplifies the calculation. But nevertheless, I need to make another substitution after, since I have the integral $$\int \frac{1}{u^2-c}du$$
I can solve it, but are there other ( perhaps shorter ) methods?
On
Firstly, the integral $$\int\frac{1}{u^2-c^2}du=-\frac{1}{c}\operatorname{artanh}\frac{u}{c}+C$$ is quite standard, so really if you'd have known it your method would be pretty efficient! :)
Secondly, you could first make the substitution $u=e^x/{\sqrt{c}}$, so that $e^x=u\sqrt{c}$ and $du/dx=e^x/\sqrt{c}=u\implies dx=du/u$. This means that if $I$ is the integral then $$\begin{align} I&=\int \frac{1}{\sqrt{e^{2x} +c}}dx\\ &=\int\frac{1}{\sqrt{u^2c+c}}\times\frac{1}{u}du\\ &=\frac{1}{\sqrt c}\int\frac{1}{u\sqrt{u^2+1}}du \end{align}$$ Here we have a not-very-standard integral: $$\int\frac{1}{u\sqrt{u^2+1}}du=-\operatorname{arcosech}u+C=-\operatorname{arsinh}\frac{1}{u}+C$$Can you finish it from there?
I hope you found that helpful. If you have any questions please don't hesitate to ask. :)
You can write this integral as $$ I = \int \frac{e^{-x}dx}{\sqrt{1+ce^{-2x}}}$$ and make the substitution (for $c>0$): $$ \sqrt{c}e^{-x} = \sinh t$$ $$ \sqrt{1+ce^{-2x}} = \cosh t$$ $$ -\sqrt{c}e^{-x}dx = \cosh t\,dt $$ then you immediately get $$ I = -\frac{1}{\sqrt{c}}\int dt = -\frac{1}{\sqrt{c}}{\rm arsinh}(\sqrt{c}e^{-x}) + const.$$ For $c<0$ there's a similar substitution, $\sqrt{-c}e^{-x} = \sin t$.