Integral with radical in denominator: $\int \frac{dx}{x(x^2-1)^{3/2}}$

Question

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.

Answer 1

Let $x=\sec\theta$, $dx=\sec\theta\tan\theta d\theta$ to get

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

Answer 2

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.

Answer 3

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

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 $t = \sin \theta$

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

Answer 4

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.

Answer 5

It is well known that the trigonometric substitution $x=\sec\theta$ is problematic if one is not careful with mapping domains, or back-substitution, to accommodate both $x>1$ and $x<-1$. It is desirable to avoid it when possible.

Substitute $t= \sqrt{x^2-1}$ instead

\begin{align} \int \frac{1}{x(x^2-1)^{3/2}}dx &= \int\frac1{t^2(t^2+1)}dt =- \frac1t - \tan^{-1}t\\ &=- \frac1{\sqrt{x^2-1}}- \tan^{-1}\sqrt{x^2-1}+C \end{align}

Answer 6

Let $x=\cosh u$. Then $$\int \frac{dx}{x(x^2-1)^{3/2}}=\int\frac{du}{\cosh u\sinh^2u}=\int\frac{\cosh u\,du}{\sinh^2 u(\sinh^2u+1)}\\=\int\frac{\cosh u\,du}{\sinh^2u}-\int\frac{\cosh u\,du}{\sinh^2u+1}=-\frac{1}{\sinh u}-\arctan\sinh u+c\\=-\frac{1}{\sqrt{x^2-1}}-\arctan(\sqrt{x^2-1})+c.$$

Answer 7

Substituting $y = \sqrt{x^2-1}-x$ gives

$$\int\frac{dx}{(x^3-x)\sqrt{x^2-1}} = \int \frac{8y^2}{(1+y^2) (1-y^2)^2} \,dy$$

and the subsequent integrand has a simple partial fraction expansion.