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I tried trigonometric substitution but it got me nowhere, and I can't find any examples online which has a radical in the denominator and a factor of $x$ outside of it.

Own attempt:

$$\int \frac{dx}{x(x^2-1)^{3/2}} = \int\frac{dx}{(x^3-x)\sqrt{x^2-1}}$$

Let $x = \sec t \iff dx = \arccos\frac{1}{t}\,dt$

$$\int\frac{\arccos\frac{1}{t}dt}{(\sec^3 t-\sec t)\tan t}$$

This is where I give up; it's more difficult because I haven't worked with $\sec$ up until now either; and it's not taught to us either.

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$$ \int \frac{dx}{x (x^2-1)^{3/2}} = \int \frac{dx}{x^4\left(1 - \frac{1}{x^2}\right)^{3/2}} $$

Substitute $t = \frac{1}{x} \Rightarrow dt = -\frac{dx}{x^2} $

$$ \int \frac{dx}{x^4\left(1 - \frac{1}{x^2}\right)^{3/2}} = -\int \frac{t^2 \, dt}{(1 - t^2)^{3/2}} $$

Now do the trig substitution $t = \sin \theta$

$$ -\int \frac{t^2 \,dt}{(1 - t^2)^{3/2}} = -\int \frac{\sin^2 \theta \cos \theta \, d\theta}{\cos^3 \theta} = -\int \tan^2 \theta\, d\theta = -\int (\sec^2 \theta - 1) \,d\theta \\= -\tan \theta + \theta + C = -\frac{t}{\sqrt{1-t^2}} + \arcsin t + C = -\frac{1}{\sqrt{x^2-1}} + \arcsin \left(\frac{1}{x}\right) + C $$

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  • $\begingroup$ In the first line, how do you factor $x^2$ out of the denominator like that? $\endgroup$ – B. Lee Nov 23 '14 at 21:28
  • $\begingroup$ $(x^2 - 1)^{3/2} = (x^2(1 - 1/x^2))^{3/2} = (x^2)^{3/2}(1 - 1/x^2)^{3/2} = x^3 (1 - 1/x^2)^{3/2}$ $\endgroup$ – Dylan Nov 23 '14 at 21:31
  • $\begingroup$ I actually made a mistake. Fixed it now. $\endgroup$ – Dylan Nov 23 '14 at 21:34
  • $\begingroup$ Hmm. I've been trying to make your solution work, taking the derivative in Wolfram gives $\frac{x}{(x^2-1)^{3/2}}-\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}$ $\endgroup$ – B. Lee Nov 23 '14 at 23:11
  • $\begingroup$ That is in fact the same function, in a different form. $\endgroup$ – Dylan Nov 25 '14 at 4:37
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Let $x=\sec\theta$, $dx=\sec\theta\tan\theta d\theta$ to get

$\displaystyle\int\frac{\sec\theta\tan\theta}{\sec\theta\tan^{3}\theta} d\theta=\int\cot^{2}\theta d\theta=\int(\csc^{2}\theta-1) d\theta$.

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  • $\begingroup$ While I really appreciate it, this didn't help me at all. I just learned a new trig function $\csc$ from your comment :S. I think I'll try substituting to get the radical in form $\sqrt{x^2+1}$ and then use tangent directly; maybe that will be easier. $\endgroup$ – B. Lee Nov 23 '14 at 20:41
  • $\begingroup$ You can find $\int\cot^{2}\theta d\theta$ by writing it as $\int\cos\theta\big(\frac{\cos\theta}{\sin^{2}\theta}\big)d\theta$ and then integrating by parts, with $u=-\cos\theta$ and $dv=-\frac{\cos\theta}{\sin^{2}\theta} d\theta$, if you want to avoid using the cosecant. $\endgroup$ – user84413 Nov 23 '14 at 20:49
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    $\begingroup$ $\cot$ and $\csc$ have the same relationship as $\tan$ and $\sec$. $\frac{d}{dx}\cot x = -\csc^2 x$ and $\csc^2 x = \cot^2 x + 1$ $\endgroup$ – Dylan Nov 23 '14 at 21:29
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If $x^2-1=t$ then $2xdx=dt$ so $dx/x=2dt/(t+1)$. So now we have $\int \frac{dt}{2(t+1)t{^{3/2}}}$.
Set $t^{1/2}=u $, then $t=u^{2}$ and $dt=2udu$ and after substituting, we eventually have $\int \frac{du}{u^{2}(u^{2}+1)}$ which is easy to compute.

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If $x=\sec(t)$, we have $dx=\sec(t)\tan(t)$. Hence, the integral becomes $$I = \int \dfrac{\sec(t)\tan(t)dt}{\sec(t)\sqrt[3/2]{\sec^2(t)-1}} = \int \dfrac{\sec(t)\tan(t)dt}{\sec(t)\tan^3(t)} = \int \cot^2(t)dt$$ Trust you can finish it off from here.

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