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What is $$\int\dfrac{x^2+5x+8}{\sqrt{x^2-8x}}\,dx\quad?$$

I've tried all of the methods I know, but they lead to extremely complex things that I don't know how to solve, so either I'm missing something simple, or I don't have the tools to solve this. I tried standard u substitution with $u=x^2-8x$ and $u=\sqrt{x^2-8x}$. I don't know how to rewrite this to attempt to use partial fractions, but I feel that's the best candidate. Also I don't think you can use trig substitution (at least, I have never seen one where "a" is not a square and has a variable with it).

Can anyone provide the first steps or method to solve this?

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  • $\begingroup$ Integral of $\int$ is double integral. Is this correct or a typo? $\endgroup$
    – UserX
    Sep 3 '14 at 21:01
  • $\begingroup$ @user148432 Based on the question, my guess would be a typo. $\endgroup$
    – qwr
    Sep 3 '14 at 21:09
  • $\begingroup$ I don't know what a double integral is, so where ever you're seeing that, it must be a typo. But out of curiosity, what indicates that this is a double integral? $\endgroup$
    – Sabien
    Sep 3 '14 at 21:30
  • $\begingroup$ @Sabien I believe the other posters are commenting on your title and first line where you say "what is the integral of $\int \dots$" so maybe to remove confusion change the title up a little..but to be honest it was clear for me. $\endgroup$
    – Chinny84
    Sep 3 '14 at 21:56
  • $\begingroup$ Oh! Haha I see now, apologies. Yes, typo indeed. $\endgroup$
    – Sabien
    Sep 3 '14 at 21:58
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First step $$ x^2-8x = \left(x-4\right)^2-16 $$ second step $$ \sqrt{x^2-8x} = 4\sqrt{u^2-1} $$ where $\dfrac{x-4}{4} = u$ should yield an integral like

$$ \int \dfrac{au^2+bu+c}{\sqrt{u^2-1}}du $$

find the a,b and c and also you can split the integral.

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  • $\begingroup$ I see, you used complete the square? How did you see to use this? Just experience, or was there something else that gave it away? $\endgroup$
    – Sabien
    Sep 3 '14 at 21:39
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    $\begingroup$ It was very much the latter for me, experience. Knowing what type of transformations will or will not work. The trick is to not be down hearted by the apparent swiftness, as behind that swiftness is a lot of pen of paper work..so you will get there! I give you an example, I am currently going through the topic of stochastic processes, and I am not so swift at that ;). Always improve your mathematical "toolbox". Good luck. $\endgroup$
    – Chinny84
    Sep 3 '14 at 21:47
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We have: $$ x^2-8x = (x-4)^2 -16 $$ hence by setting $x=4y+4$ we have: $$I=\int\frac{x^2+5x+8}{\sqrt{x^2-8x}}\,dx = \int\frac{16y^2+52y+44}{\sqrt{y^2-1}}dy$$ and now, by setting $y=\cosh z$ we have: $$ I = \int(16\cosh^2 z+52\cosh z + 44)\,dz = \int (8\cosh(2z)+52\cosh z+52)\,dz \\= 4\sinh(2z)+52\sinh z+52z.$$

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All such integrals can be evaluated by an Euler substitution. In this case let $$t=x-\sqrt{x^2-8x}$$ solve for $x$. In then end you will obtain a rational function of $t$.

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$\displaystyle\int\frac{x^2+5x+8}{\sqrt{x^2-8x}} dx=\int\frac{x^2+5x+8}{\sqrt{(x-4)^2-16}} dx$.

Now let $x-4=4\sec\theta$, $x=4(\sec\theta+1)$, $dx=4\sec\theta\tan\theta d\theta$ to get

$\displaystyle\int\frac{16(\sec\theta+1)^2+5(4)(\sec\theta+1)+8}{4\tan\theta}4\sec\theta\tan\theta d\theta$

$\displaystyle=\int(16\sec^{3}\theta+52\sec^{2}\theta+44\sec\theta) d\theta$

$\displaystyle=16\left[\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)\right]+52\tan\theta+44\ln|\sec\theta+\tan\theta|+C$

$\displaystyle=8\sec\theta\tan\theta+52\ln|\sec\theta+\tan\theta|+52\tan\theta+C$

$=\displaystyle\frac{1}{2}(x-4)\sqrt{x^2-8x}+52\ln\lvert x-4+\sqrt{x^2-8x}\rvert+13\sqrt{x^2-8x}+C$

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