rigliztetbf

2022-06-24

How to integrate $\int \frac{dx}{x\sqrt{{x}^{4}-1}}$

Ryan Newman

Expert

So if you write
${\mathrm{tan}}^{-1}\sqrt{{x}^{4}-1}=\theta ,$
Then
$\mathrm{tan}\theta =\sqrt{{x}^{4}-1}.$
A right triangle that tells this story has $\theta$ as one angle, $\sqrt{{x}^{4}-1}$ as the opposite side and 1 as the adjacent side. Using Pythagorean theorem we can work out the length c of the hypotenuse:
$\left(\sqrt{{x}^{4}-1}{\right)}^{2}+{1}^{2}={c}^{2}$
which shows that $c={x}^{2}$
So $\mathrm{sec}\theta ={x}^{2}/1$, that is ${\mathrm{sec}}^{-1}\left({x}^{2}\right)=\theta ={\mathrm{tan}}^{-1}\left(\sqrt{{x}^{4}-1}\right).$

Zion Wheeler

Expert

We have,
$\int \frac{dx}{x\sqrt{{x}^{4}-1}}$
Let ${x}^{2}=\mathrm{sec}\left(\theta \right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}2x\phantom{\rule{thinmathspace}{0ex}}dx=\mathrm{sec}\left(\theta \right)\mathrm{tan}\left(\theta \right)\phantom{\rule{thinmathspace}{0ex}}d\theta$
Which further implies that, $dx=\frac{\sqrt{\mathrm{sec}\left(\theta \right)}\mathrm{tan}\left(\theta \right)}{2}\phantom{\rule{thinmathspace}{0ex}}d\theta$
After substitution, the given integral changes to:
$\frac{1}{2}\int \frac{\sqrt{\mathrm{sec}\left(\theta \right)}\mathrm{tan}\left(\theta \right)}{\sqrt{\mathrm{sec}\left(\theta \right)}\cdot \sqrt{{\mathrm{sec}}^{2}\left(\theta \right)-1}}\phantom{\rule{thinmathspace}{0ex}}d\theta$
$=\frac{1}{2}\int \frac{\sqrt{\mathrm{sec}\left(\theta \right)}\mathrm{tan}\left(\theta \right)}{\sqrt{\mathrm{sec}\left(\theta \right)}\cdot \sqrt{{\mathrm{tan}}^{2}\left(\theta \right)}}\phantom{\rule{thinmathspace}{0ex}}d\theta$
$=\frac{1}{2}\int \frac{\sqrt{\mathrm{sec}\left(\theta \right)}\mathrm{tan}\left(\theta \right)}{\sqrt{\mathrm{sec}\left(\theta \right)}\mathrm{tan}\left(\theta \right)}\phantom{\rule{thinmathspace}{0ex}}d\theta$
$=\frac{1}{2}\int d\theta$
$=\frac{1}{2}\theta +C$
$=\overline{)\frac{1}{2}{\mathrm{sec}}^{-1}\left({x}^{2}\right)+C}$

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