Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$
$z = 4\sec\theta$
$dz = 4\sec\theta \tan d\theta$
$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$
$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$
I am again stuck, I have no idea how to proceed.
 A: Hint:
$$\frac{1}{\sec^2(\theta)}=\cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))$$
$$z=4\sec(\theta)$$
$$2\theta=2\sec^{-1}(\frac{z}{4})$$
$$\sin(2\theta)=\sin(2\sec^{-1}(\frac{z}{4}))=\frac{8\sqrt{z^2-16}}{z^2}$$
A: First step. Erase that trigonometric substitution. Then you can make $\sqrt{z^2-4}=z+t$, which is the first Euler's substitution. 
the second Euler substitution also work. This would mean to make $\sqrt{z^2-4}=(z-2)t$.
A: Putting $z=2\sec\theta, dz=2\sec\theta\tan\theta d\theta$
$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{2\sec\theta\tan\theta d\theta}{(2\sec\theta)^32\tan\theta}=\frac18\int\cos^2\theta d \theta=\frac1{16}\int(1+\cos2\theta)d\theta$$
$$=\frac1{16}\left(\theta+\frac{\sin2\theta}2\right)+K$$
Now, $\cos\theta =\frac2z\implies \theta =\arccos \frac2z $
$\implies \sin\theta=\sqrt{1-\left(\frac2z\right)^2}=\frac{\sqrt{z^2-4}}z$
$\implies  \sin2\theta=2\sin\theta\cos\theta=2\cdot \frac{\sqrt{z^2-4}}z\cdot \frac2z$
A: Putting $z^2-4=u^2, zdz=udu$
$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{udu}{(u^2+4)^2u}=\frac{du}{(u^2+2^2)^2}$$
This can be addressed using Trigonometric identities $u=2\tan\theta$ like my other answer  
Alternatively, $$\text{Let }I_n=\int \frac {du}{(u^2+a^2)^n}$$
$$=\int du\cdot\frac1{(u^2+a^2)^n}-\int\left(\int du \frac{d\frac1{(u^2+a^2)^n}}{du}\right)du  $$
$$=\frac u{(u^2+a^2)^n}+2n\int \frac{u^2}{(u^2+a^2)^{n+1}}du$$
$$=\frac u{(u^2+a^2)^n}+2n\left(\int \frac1{(u^2+a^2)^n}-a^2\int \frac1{(u^2+a^2)^{n+1}} du\right)$$
$$\implies I_n=\frac u{(u^2+a^2)^n}+2n\left(I_n-a^2I_{n+1}\right)$$
$$\implies 2a^2n\cdot I_{n+1}=\frac u{(u^2+a^2)^n}+(2n-1)I_n $$
$$n=1\implies 2a^2  \int \frac {du}{(u^2+a^2)^2}=\frac u{(u^2+a^2)}+\int \frac {du}{(u^2+a^2)}=\frac u{(u^2+a^2)}+\frac1a\cdot \arctan \frac ua+K$$
$$a=2\implies 8  \int \frac {du}{(u^2+4)^2}=\frac u{(u^2+4)}+\int \frac {du}{(u^2+4)}=\frac u{(u^2+4)}+\frac12\cdot\arctan \frac u2+K$$
Now, $u=\sqrt{z^2-4}, u^2+4=z^2$
