How to integrate $ \int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx$? How to integrate $$ \int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx\,\,?$$
The given answer is $$ \color{brown}I=-\frac{1}{\sqrt{33}}\cdot \tan^{-1}\bigg(\frac{\sqrt{3x^2-6x-3}}{\sqrt{11}\cdot (x-3)}\bigg)+\mathcal{C}.$$
I tried by different substitutions i.e $\frac{x^2 - 2x -1}{x-3} = t$, but I am not getting my desired answer.
$ORIGINAL$ $QUESTION$:
This question was asked in our test and the given answer was option D ,i.e none on the given options were correct.

$$I=\int \frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}\,dx$$ This can be simplifies using
$$\frac{x-2}{7x^2-36x+48}=\frac 1{7(a-b)}\left(\frac{a-2 } {x-a }+\frac{2-b } {x-b } \right)$$ where
$$a=\frac{2}{7} \left(9-i \sqrt{3}\right) \qquad \text{and} \qquad b=\frac{2}{7} \left(9+i \sqrt{3}\right) $$ which makes that we are facing two integrals
$$I_c=\int \frac {dx} {(x-c)\sqrt{x^2-2x-1}}$$
Complete the square and let $x=1+\sqrt 2 \sec(t)$ which gives
$$I_c=\int \frac{dt}{(1-c) \cos (t)+\sqrt{2}}$$ Now, using the tangent half-angle subtitution
$$I_c=2\int\frac{du}{\left(c+\sqrt{2}-1\right) u^2-c+\sqrt{2}+1}=\frac{2 }{\sqrt{-c^2+2 c+1}}\tan ^{-1}\left(u\frac{\sqrt{c+\sqrt{2}-1}
   }{\sqrt{-c+\sqrt{2}+1}}\right)$$ and so on ....
