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.

$ORIGINAL$ $QUESTION$:enter image description here

This question was asked in our test and the given answer was option D ,i.e none on the given options were correct.

  • 1
    $\begingroup$ This integral exceeds the standard computation time for WolframAlpha :) (The differential of the answer does not, though.) $\endgroup$ Jul 13, 2020 at 4:57
  • $\begingroup$ I am a JEE aspirant and this question was asked in our advanced test, so I think there might be a way to solve this problem using fundamentals of integration,@AniruddhaDeb $\endgroup$
    – Anonymous
    Jul 13, 2020 at 5:01
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    $\begingroup$ The answer can be rewritten as $$-\frac{1}{3\sqrt{\frac{11}{3}}} \cdot \arctan \left( \frac{\sqrt{x^2 - 2x -1}}{(x-3)\sqrt{\frac{11}{3}}} \right)$$, which suggests that a possible substitution is $\frac{x^2 - 2x -1}{x-3} = t$. I can't see any other patterns here unfortunately. Also, if it was asked in a test, it must have been an MCQ or somewhat similar question, so reverse engineering would be a viable solution given the limited time you have in the test. $\endgroup$ Jul 13, 2020 at 5:07
  • $\begingroup$ yes I tried taking $\frac{x^2 - 2x -1}{x-3} = t$ but endedup with an undesired answer,@AniruddhaDeb $\endgroup$
    – Anonymous
    Jul 13, 2020 at 5:09
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    $\begingroup$ This is not an integration question; it's a differentiation question and the answer is D because when you differentiate $arctan$ or some rational expression containing a single square root, any square root in the result must be a multiple of the square root you started with. $\endgroup$
    – Misha
    Jul 13, 2020 at 5:24

3 Answers 3


$$I=\int \frac{x-2}{(7x^2-36x+48)\sqrt{x^2-2x-1}}\,dx$$ 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 ....

  • $\begingroup$ Thank you @Claude Leibovici $\endgroup$
    – Anonymous
    Jul 13, 2020 at 6:36
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    $\begingroup$ @VenkatAmith. Quite tedious, isn't it ? But doable. $\endgroup$ Jul 13, 2020 at 7:05
  • $\begingroup$ yes, @Claude Leibovici, you re correct it is tedious and doable, but I had not to encounter this problem before! $\endgroup$
    – Anonymous
    Jul 13, 2020 at 7:11
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    $\begingroup$ @VenkatAmith. Me neither ! $\endgroup$ Jul 13, 2020 at 7:12

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.

  • $\begingroup$ Thank you @Nikunj $\endgroup$
    – Anonymous
    Jul 13, 2020 at 6:28
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    $\begingroup$ You're welcome and all the best for advanced! :) $\endgroup$
    – Nikunj
    Jul 13, 2020 at 6:29
  • $\begingroup$ thanks, I thought that this could be done through the applications or substitutions used in indefinite integration which we leart in JEE syllabus, but this doesn't seem like that, but no worries, thanks a lot @Nikunj $\endgroup$
    – Anonymous
    Jul 13, 2020 at 6:33

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.


Differentiation yields the following expression, comparing it with the integrand gives equations in $a$, $b$ and $c$, solving which gives the antiderivative.


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 $\arctan()$.

  • $\begingroup$ Thank you @Paras Khosla $\endgroup$
    – Anonymous
    Jul 13, 2020 at 9:15

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