Semaj Christian

2022-06-19

What is going on with this integral $\int \frac{dx}{\sqrt{{e}^{2x}-9}}$

Myla Pierce

Rewrite the integrand by multiplying and dividing by ${e}^{x}$
$\int \frac{{e}^{x}dx}{{e}^{x}\sqrt{{e}^{x}-9}}=\int \frac{d\left({e}^{x}\right)}{{e}^{x}\sqrt{{e}^{x}-9}}=\frac{1}{3}{\mathrm{sec}}^{-1}\left(\frac{{e}^{x}}{3}\right)+C$
By drawing a triangle we can see that
$\mathrm{sec}3\theta =\frac{{e}^{x}}{3}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}3\theta =\frac{\sqrt{{e}^{2x}-9}}{3}$
Thus we also obtain your arctan solution, which is equivalent to the arcsec solution
$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{\sqrt{{e}^{2x}-9}}{3}\right)+C$
and similarly, we can obtain the arcsin solution
$\mathrm{sin}3\theta =\frac{{e}^{x}}{\sqrt{{e}^{2x}-9}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\mathrm{arcsin}\left(\frac{1}{\sqrt{1-9{e}^{-x}}}\right)+C$

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