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.

Remember:
For such questions, It is always better to differentiate the options in an MCQ format exam.
Since you'd like an approach to integrate this, here goes:)
$$\int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}\, dx = \int\frac{(x-1-1)\,dx}{(7x(x-1)-29(x-1)+19)\sqrt{(x-1)^2-2}}$$
Put $x-1 \rightarrow v$,
$$ = \int\frac{v-1}{(7v^2-22v+19)\sqrt{v^2-2}}\, dv$$
Now put $v \rightarrow \sqrt2\sec y$, $dv = \sqrt 2 \sec y\tan y\,dy$
$$ = \int\frac{(\sqrt2\sec y-1)\sec y}{14\sec^2y - 22\sqrt2\sec y+ 19}\,dy$$
$$ = \frac{1}{7\sqrt2}\int\frac{\sec^2y - (11\sqrt2/7)\sec y + 19/14 + (15\sqrt2/7)\sec y-19/14}{\sec^2y - (11\sqrt2/7)\sec y+ 19/14}\,dy$$
$$ = \frac {y}{7\sqrt2} + \frac{15}{49}\int\frac{\sec y - 19/(30\sqrt2)}{\sec^2y - (11\sqrt2/7)\sec y+ 19/14}\, dy$$
Our denominator here doesn't have real roots. The way ahead would be to factorize the denominator as $(\sec y - a)(\sec y - b)$ where $a$ and $b$ are the complex and then use partial fractions.
