TITLE: How to integrate $ \int\frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}dx$?
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: By inspection and from the options too, it is clear that the antiderivative has to be of the following form with constants $a$, $b$ and $c$ remaining to be determined.
$$c\arctan\left(\frac{\sqrt{x^2-2x-1}}{ax+b}\right)$$
Differentiation yields the following expression, comparing it with the integrand gives equations in $a$, $b$ and $c$, solving which gives the antiderivative.
$$\frac{c\left(\left(b+a\right)x-b+a\right)}{\sqrt{x^2-2x-1}\left(\left(a^2+1\right)x^2+\left(2ab-2\right)x+b^2-1\right)}$$
But obviously the answer had to be something different from the ones given in the options, because the term $\sqrt{x^2-2x-1}$ has to remain put inside the $\arctan()$.