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I'm reviewing for an intro calculus exam, and the following problem appears on a past final exam:

If: $$\int^3_1 \frac{1}{x^4\sqrt{1+x}}\, dx = k$$

What is: $$\int^3_1 \frac{1}{x^5\sqrt{1+x}}\, dx $$ (I'm assuming the answer will be in terms of $k$)

It seems that most basic integration techniques(substitution, integration by parts, trig sub, etc.) will not allow the solution of the integral, and I'm not sure how else to approach this problem at my level. I've run this by both my lecturers, and they cannot find a solution in a reasonable amount of time either. I'm curious because it seems the there would be a simple solution or rule I'm ignorant of (considering this is on an intro calc exam), but I'm stumped. Where am I going wrong? Thanks!

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    $\begingroup$ I have a solution, but I don't have the room to write it in this margin! But actually I'm on mobile and don't want to write integrals. Best I can do now is a spiritual guide version. Do parts on the first equation, with $1/x^4$ as one of the terms. Then you get some square roots in the numerator. Multiply by that on top and bottom, and split the fraction into its two parts. Magic occurs. $\endgroup$
    – Zach Stone
    Aug 12 '15 at 4:53
  • $\begingroup$ @ZachStone: Awesome if true! $\endgroup$
    – Brian Tung
    Aug 12 '15 at 4:53
  • $\begingroup$ @ZachStone What is this witchcraft, it worked! I don't know how I missed this solution. Thanks for the help! $\endgroup$
    – Ian Shi
    Aug 12 '15 at 5:03
  • $\begingroup$ @IanShi I'm just glad that you could follow my explanation. This is my new favorite hard integral. $\endgroup$
    – Zach Stone
    Aug 12 '15 at 5:05
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    $\begingroup$ Being "elderly" at this point, I looked at the integrand and thought, "I'll bet this is in Gradshteyn & Ryzhik." The relevant indefinite integrals are in $ \ 2.228 \ $ and $ \ 2.224 \ $ (related). For $ \ a = b = 1 \ $ , it shows $$ \ \int \ \frac{dx}{x^2 \ \sqrt{1+x}} \ = \ -\frac{\sqrt{1+x}}{x} -\frac{1}{2} \int \ \frac{dx}{x \ \sqrt{1+x}} \ $$ and $$ \ \int \ \frac{dx}{x^3 \ \sqrt{1+x}} \ = \ ( \ -\frac{1}{2x^2} + \frac{3}{4x} \ ){\sqrt{1+x}}+\frac{3}{8} \int \ \frac{dx}{x \ \sqrt{1+x}} \ \ . $$ So the general "reduction formula" for this family appears to be pretty convenient. $\endgroup$ Aug 12 '15 at 6:19
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Let $I_4$ be the given integral (i.e. k) and let $I_5$ be the integral you want. I am just expanding on Zach Stone's comment, so credit due to him, one can intgrate by parts and write

$$I_4 = \int^3_1 \frac{1}{x^4}\frac{1}{\sqrt{1+x}}\, dx $$ which will give $$I_4 = \left. \frac{2\sqrt{1+x}}{x^4}\right|_1^3 + \int^3_1 \frac{8\sqrt{1+x}}{x^5}dx$$

The trick is now to multiply and divide the second term within the integral by $\sqrt{1+x}$. The equation then easily simplifies to

$$I_4 = \frac{4}{81} - 2\sqrt{2} +8I_4 +8I_5$$ which gives the required relation between $I_4$ and $I_5$. I verified that it agrees with Brian Tung's calculations from Wolfram.

So, in short, yes, this was possible using simple integration by parts. Once again, credit to Zach Stone

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    $\begingroup$ Looks good! Thanks for doing the dirty work of texing everything. $\endgroup$
    – Zach Stone
    Aug 12 '15 at 6:04
  • $\begingroup$ I learnt a lot during that. I try not to answer questions which require that. I need to get over that roadblock.. $\endgroup$
    – Shailesh
    Aug 12 '15 at 6:05
  • $\begingroup$ @user103828 Thanks for the edits and removing the kinks. $\endgroup$
    – Shailesh
    Aug 12 '15 at 6:19
  • $\begingroup$ Applause to Shailesh and ZachStone. +1 $\endgroup$
    – Brian Tung
    Aug 12 '15 at 20:29
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Weird. WolframAlpha gives

$$ k = \frac{351\sqrt{2}-226+405\tanh^{-1}2-405\tanh^{-1}\sqrt{2}}{648} \doteq 0.20972 $$

and the other integral has value

$$ \frac{1550-1161\sqrt{2}+2835\tanh^{-1}2-2835\tanh^{-1}\sqrt{2}}{5184} \doteq 0.16387 $$

There's no obvious exact relationship that I can see between those. It's pretty clear that the second integral must be smaller than the first, and I'm sure I could obtain some tighter bounds than that, but I don't know how I would obtain an exact expression in terms only of $k$.

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Adding to Brian Tung's answer let $I_{4}$ be the integral with $x$ to the $4^{th}$ power and $I_{5}$ be the integral in question. It can be determined that $$I_{5} = - \frac{7 \, k}{8} + \frac{1}{96} \, \left[ 116 - 67 \, \sqrt{2} - 210 \, \tanh^{-1}(2) \right].$$ It should be noted that $\tanh^{-1}(2)$ and $\tanh^{-1}(\sqrt{2})$ are complex in nature.

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I think this is just a roundabout version of Zach's solution, but I also think it might be the intended solution considering how nicely the bounds work out, so I thought it was worth sharing.

Substituting $u=\sqrt{1+x}$ turns these integrals into

$$2\int_{\sqrt{2}}^2 \frac{1}{(u^2-1)^4}\, du=k\quad\text{and}\quad2\int_{\sqrt{2}}^2 \frac{1}{(u^2-1)^5}\, du=I.$$

We want to find a relationship between $I$ and $k$.

Now, let $u=\sec\theta$ in the each integral, so $du=\sec\theta\tan\theta$, and we have

$$k=2\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot^8\theta\sec\theta\tan\theta\, d\theta\quad\text{and}\quad I=2\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot^{10}\theta\sec\theta\tan\theta\, d\theta.$$

Now, integrate by parts, with $u=\cot^{8}\theta$, so $du=-8\cot^7\theta\csc^2\theta$ and $dv=\sec\theta\tan\theta$, so $v=\sec\theta$. Then the integral is

$$2\left[\sec\theta\cot^8\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}+16\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^7\theta\csc^2\theta\sec\theta\, d\theta=\frac{4}{81}-2\sqrt{2}+16\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^7\theta\sec\theta(1+\cot^2\theta)\, d\theta.$$

Focusing on the integral, we can break it up into

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^7\sec\theta\, d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^9\theta\sec\theta\,d\theta=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^8\theta\sec\theta\tan\theta\,d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot^{10}\theta\sec\theta\tan\theta.$$

But this is just $I+k$, so we have

$$k=\frac{2}{81}-\sqrt{2}+8I+8k\implies I=\frac{1}{8}\left(2\sqrt{2}-\frac{4}{81}-7k\right).$$

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  • $\begingroup$ Aside: your substitution could have been discovered by wanting to use $x = \tan^2 \theta$ to convert the square root into something that can be simplified. Also, normalizing $$\cot^n \theta \sec \theta \tan \theta = \frac{\cos^{n-2} \theta}{\sin^{n-1} \theta} = \cot^{n-2} \theta \csc \theta$$ may be simpler to work with. $\endgroup$
    – user14972
    Aug 12 '15 at 8:33
  • $\begingroup$ My main reason for leaving it in that form was to make the integration by parts more straightforward. $\endgroup$
    – Samir Khan
    Aug 12 '15 at 23:17

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