 Caleb Proctor

2022-07-04

Antiderivatives such as $\int \frac{dx}{\sqrt{{e}^{2x}+c}}$ Jamarcus Shields

Expert

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 $du/dx={e}^{x}/\sqrt{c}=u\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}dx=du/u$. This means that if I is the integral then $\begin{array}{rl}I& =\int \frac{1}{\sqrt{{e}^{2x}+c}}dx\\ & =\int \frac{1}{\sqrt{{u}^{2}c+c}}×\frac{1}{u}du\\ & =\frac{1}{\sqrt{c}}\int \frac{1}{u\sqrt{{u}^{2}+1}}du\end{array}$
Here we have a not-very-standard integral: $\int \frac{1}{u\sqrt{{u}^{2}+1}}du=-\mathrm{arcosech}u+C=-\mathrm{arsinh}\frac{1}{u}+C$ ntaraxq

Expert

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

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