How to solve the integral
$$\int \frac{1}{\left(x-2\right)^3\sqrt{3x^2-8x+5}}dx$$
I am pretty sure that a certain substitution should work. I tried using $x-2=\frac{1}{t}$ and got $$\int \:\frac{-\frac{dt}{t^2}}{\frac{1}{t^3}\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
After some simplification: $$\int \:\frac{tdt}{\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
Any this point, I'm lost.
From the point that you stopped:
$$=\int\frac{tdt}{\sqrt{5-5\left(\frac 1 t+2\right)^2}}=\int\frac{tdt}{\sqrt{-\frac{5}{t^2}-\frac{20}{t}-15}}=-\frac{i}{\sqrt 5}\int\frac{t^2}{\sqrt{3t^2+4t+1}}dt$$
$$\overset{\text{complete the square }}{=}-\frac{i}{\sqrt 5}\int\frac{t^2}{\sqrt{\left(\sqrt 3 t+\frac{2}{\sqrt 3}\right)^2-\frac 1 3}}dt$$
Set $u=\sqrt t+\frac{2}{\sqrt 3}$ and $du=\sqrt 3 dt$
$$=-\frac{i}{\sqrt 5}\int\frac{(2\sqrt 3-3u)^2}{27\sqrt{u^2-\frac 1 3}}du$$
substitute $u=\frac{\sec s}{\sqrt 3}$ and $du=\frac{\tan s \sec s}{\sqrt 3}ds$ then $\sqrt{u^2-\frac 1 3}=\sqrt{\frac{\sec^2 s}{3}-\frac 1 3}=\frac{\tan s}{\sqrt 3}$ and $s=\sec^{-1}(\sqrt 3 u)$
$$=-\frac{i}{81\sqrt 5}\int \sqrt 3 \sec s(2\sqrt 3-\sqrt 3\sec s)^2$$
$$=-\frac{i}{9\sqrt{15}}\int \sec^3 s ds+\frac{4i}{9\sqrt{15}}\int \sec^2 s ds-\frac{4i}{9\sqrt{15}}\int \sec s ds=\dots$$