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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 $

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    @DavidMitra I understand to use $dx\over {x}$ but where do you get $dx\over {x}$ = $du\over {2u}$ rather than equals $du\over {2x}$?2012-04-06

1 Answers 1

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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.

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    +1. Sorry for being a bit stubborn, thanks for the help David.2012-04-06