Clever way of calculating the integral $ \int \frac{dt}{t^2\sqrt{t-2} } $ 
$$ \int  \frac{\text{d}t}{t^2\sqrt{t-2} } $$

I know it can be calculated using somewhat complicated substitutions, but is there possibly some clever way of solving that type of integral? I don't even expect full solution, just ideas.
 A: Substitute $t=2\tan^2(u)+2$:
$$
\begin{align}
\int\frac{\mathrm{d}t}{t^2\sqrt{t-2}}
&=\frac1{\sqrt2}\int\cos^2(u)\,\mathrm{d}u\\
&=\frac1{2\sqrt2}\int(1+\cos(2u))\,\mathrm{d}u\\
&=\frac1{2\sqrt2}\left(u+\frac12\sin(2u)\right)+C\\
&=\frac1{2\sqrt2}\left(u+\frac{\tan(u)}{1+\tan^2(u)}\right)+C\\
&=\frac1{2\sqrt2}\arctan\left(\sqrt{\frac{t-2}2}\right)+\frac12\frac{\sqrt{t-2}}t+C\\
\end{align}
$$
A: Set $u=t-2$ and $du=dt$
$$=\int\frac{du}{\sqrt u (u+2)^2}$$
Set $s=\sqrt u$ and $ds=\frac{du}{2\sqrt u}$
$$=2\int\frac{ds}{ (s^2+2)^2}$$
Set $s=\sqrt 2 \tan (p)$ so $p=\arctan (s/\sqrt 2)$
$$=\frac{1}{\sqrt 2}\int\cos^2(p)dp$$
Write as
$$=\frac{1}{\sqrt 2}\int \bigg(\frac 1 2\cos(2p)+\frac 1 2 \bigg)dp$$
$$=\frac{p}{2\sqrt 2}+\frac{\sin(2p)}{4\sqrt 2}+C$$
$$ \boxed{=\frac{2\sqrt{t-2} + \sqrt{2}t \arctan\left(\sqrt{\frac{t-2}{2}}\right)}{4t}+C}$$
A: HInts
1. use $x = t-2$, $\text{d}x = \text{d}t$
2. You're now with $\int \frac{\text{d}x}{(x+2)^2\sqrt{x}}$
3. Now use $y = \sqrt{x}$, $\text{d}y = \frac{1}{2\sqrt{x}}\ \text{d}x$
4. You now get $2 \int \frac{1}{(y^2 + 2)^2}\ \text{d}y$
5. Use now $y = \sqrt{2} \tan(z)$, $\text{d}y = \sqrt{2}\sec^2(z)\ \text{d}z$ and arranging...
6. ...you obtain $\frac{1}{\sqrt{2}}\int \cos^2(z)\ \text{d}z$ which is trivial
7. Final Result: 
$$\int \frac{\text{d} t}{t^2\sqrt{t-2}} = \frac{2\sqrt{t-2} + \sqrt{2}t \arctan\left(\sqrt{\frac{t-2}{2}}\right)}{4t}$$
A: \begin{align}
u & = \sqrt{t-2} \\[8pt]
u^2 & = t-2 \\[8pt]
u^2 + 2 & = t \\[8pt]
2u\,du & = dt
\end{align}
\begin{align}
\int \frac{dt}{t^2\sqrt{t-2}} & = \int \frac{2u\,du}{(u^2+2)^2 u} = \int\frac{2\,du}{(u^2+2)^2}
\end{align}
Then let $u = \sqrt 2 \tan \theta$, $du= \sqrt 2 \sec^2\theta\,d\theta$, $u^2+2=2\sec^2\theta$.  The integral becomes
$$
\int \frac{2\sqrt 2 \sec^2\theta\,d\theta}{\sec^4\theta} = 2\sqrt2 \int\cos^2\theta\,d\theta, \quad\text{etc.} 
$$
