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MathematicsEvaluating $\int \frac{1}{x\sqrt{x^4-4}} dx$
[+6] [2] Joe
[2012-04-06 03:20:08]
[ calculus integration indefinite-integrals ]
[ https://math.stackexchange.com/questions/128592/evaluating-int-frac1x-sqrtx4-4-dx ]

I am having trouble evaluating $$\int \dfrac{1}{x\sqrt{x^4-4}} dx$$

I tried making $a = 2$, $u = x^{2}$, $du = 2x dx$ and rewriting the integral as: $$\dfrac{1}{2} \int \dfrac{du}{\sqrt{u^2-a^2}} $$ But I believe something is not right at this step (perhaps when changing from $dx$ to $du$)?

I end up with: $${1\over 4} \operatorname{arcsec} \dfrac{1}{2}x^{2} + C$$

Any help would be appreciated, I feel I am only making a simple mistake. Also, for some reason, on WA, it is showing an answer involving $\tan^{-1}$ but I do not see an $a^{2} + u^{2}$ possibility. Note that I do know how sometimes (different) inverse trig functions when integrated are equal.

Ex: $$\int \dfrac{1}{\sqrt{e^{2x}-1}} dx = \arctan{\sqrt{e^{2x}-1}} + C = \operatorname{arcsec}(e^{x}) + C $$

wolframalpha.com/input/…. Press: Show Steps. - Elchanan Solomon
As noted in my original post, I already checked WA. I was looking for an easier way since there should be a direct substitution with an $a$ and a $u$, rather than jumping through hoops with 3-4 substitutions. Correct me if I'm wrong though. - Joe
Your first substitution is not correct. You have $du=2x\,dx$, but you want to replace $dx\over x$. Use ${dx\over x} = {du\over 2 u}$. This gives $\int {du\over u\sqrt {u^2-4}}$; which is an $\rm arcsec$ form. - David Mitra
@DavidMitra I understand to use $dx\over {x}$ but where do you get $dx\over {x}$ = $du\over {2u}$ rather than equals $du\over {2x}$? - Joe
[+8] [2012-04-06 03:47:11] David Mitra [ACCEPTED]

You did not make the substitution correctly (your substitution would work as you wrote it if $x$ were originally upstairs).

But the choice you made for $u$ will work:

You have $u=x^2$ and $du=2x\,dx$.

From $du=2x\,dx$, you have, dividing both sides by $2x^2$ $$\tag{1}{du\over 2x^2}={x\,dx\over x^2}.$$ Substituting $u=x^2$ on the left hand side of $(1)$ and simplifying the right hand side, we have $$ \color{maroon}{{du\over 2 u}}=\color{maroon}{{dx\over x}}.$$ Substituting into the integral gives $$\int {\color{maroon}{dx}\over\color{maroon} x \sqrt{ x^4-4}}= \int {\color{maroon}{du}\over\color{maroon}{ 2u}\sqrt {u^2-4}} $$ which is an $\rm arcsec$ form.


What would be your final answer then? It seems it would be the same as mine. I am still not entirely sure how you jumped from $du = 2x dx$ to the line below it with $du \over{2x^2}.$ Mind elaborating? I see the x cancel out on the RHS and the substitution of u in the line below it - it's the line above "Substituting into the integral" that is still puzzling me. - Joe
What jumped out at you to divide both sides by $2x^{2}?$ To try and get a $dx\over {x}$ term on the RHS? - Joe
@jay Yes, we needed to write $dx\over x$ in terms of $u$. - David Mitra
(1) When I differentiate $${1\over 4} \operatorname{arcsec} \dfrac{1}{2}x^{2} + C$$ I get the original, I must have just been careless in the beginning by forgetting the $u$ term on the bottom of the inside. - Joe
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[0] [2023-10-28 05:46:13] pie

$$I:=\int \frac{dx}{x \sqrt{x^4 -4}} $$

let $x=\sqrt{2} y$ then $dx =\sqrt{x} dy$ $$I=\frac{1}{2}\int \frac{dy}{y \sqrt{y^4 -1}} $$

let $y=\frac{1}{t} $ $dy =-\frac{dt}{t^2}$ notice $t =\frac{\sqrt{2}}{x}$ $$I=\frac{-1}{2}\int \frac{t dt}{t^2 \sqrt{\frac{1}{t^4} -1}} =\frac{-1}{2}\int \frac{t dt}{ \sqrt{1 -t ^4}} $$

let $t^2 =\sin(u)$ $2t dt =\cos(u)du$

$$I =\frac{-1}{4}u+C=\frac{-1}{4}\arcsin(t^2)+C =\frac{-1}{4} \arcsin( \frac{2}{x^2})+C $$


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