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$$ I=\int \frac{x^9}{(x^2+4)^6}\mathrm{d}x $$ Yeah I know, I can substitute: $$t=x^2+4\text{ or }2\tan\theta$$ So that: $$I=\frac12\int\frac{(t-4)^4}{t^6}\mathrm{d}t\text{ or } I=2^{-2}\int\tan^9\theta\cos^{10}\theta\mathrm{d}\theta$$ Both of which are a long tedious way* to solve, is there any easier method?

Update: I am not asking among these two, I am asking any "substitution" except these two, which is shorter.

Edit: I am very sorry I missed the ^$6$ in question.


  • Use Binomial Theorem for first and then divide by $t^6$, then integrate term wise.

  • $\require{cancel}\cancel{\text{Use Reduction formula for second or integration by parts.}}$


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    $\begingroup$ The one with $t$ is quick. $\endgroup$ – André Nicolas Aug 22 '14 at 19:10
  • $\begingroup$ @AndréNicolas I am not asking among these two, I am asking any "substitution" except these two, which is shorter. $\endgroup$ – RE60K Aug 22 '14 at 19:13
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    $\begingroup$ You seem to be given answers: In this case it is usually easier to take derivatives of all the answers to check whether they match the integrand. $\endgroup$ – Fabian Aug 22 '14 at 19:50
  • $\begingroup$ @achillehui It would be helpful if you'd mention details about this in your answer[not necessary] $\endgroup$ – RE60K Aug 22 '14 at 20:25
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Hint :

Rewrite the integrand as $$\frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x.$$


Edit :

The OP was changed into $$ \int\frac{x^9}{(x^2+4)^6}\ dx. $$ Use $x=2\tan\theta\ \Rightarrow\ dx=2\sec^2\theta\ d\theta$. $$ \int\frac{x^9}{(x^2+4)^6}\ dx=\frac14\int\tan^9\theta\ \cos^{10}\theta\ d\theta=\frac14\int\sin^9\theta\cos\theta\ d\theta. $$ Now, set $t=\sin\theta\ \Rightarrow\ dt=\cos\theta\ d\theta$. $$ \int\frac{x^9}{(x^2+4)^6}\ dx=\frac14\int t^9 dt=\frac1{40}t^{10}+C, $$ where $\sin\theta=\dfrac x{\sqrt{x^2+4}}$. The answer is $\color{blue}{\text{D}}$.

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    $\begingroup$ Darn, beat me! :) +1 $\endgroup$ – Ahaan S. Rungta Aug 22 '14 at 19:14
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    $\begingroup$ @AhaanRungta +1 too Brilliant buddy :) $\endgroup$ – Tunk-Fey Aug 22 '14 at 19:16
  • $\begingroup$ sorry see the edit $\endgroup$ – RE60K Aug 22 '14 at 19:16
  • $\begingroup$ I feel very sorry, hope my upvote compensates $\endgroup$ – RE60K Aug 22 '14 at 19:18
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    $\begingroup$ @Aditya It's okay; no need to be sorry! To compensate, you should probably give me a +500 bounty. Haha, just kidding. =P $\endgroup$ – Ahaan S. Rungta Aug 22 '14 at 19:21
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You don't have to use a substitution quite yet. If you long-divide the integrand, you get $$\frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x.$$Now, it is quite simple if the integrate term-by-term. Finishing things off: $$ \begin {align*} \displaystyle\int \frac {x^9}{x^2+4} \, \mathrm{d}x &= \displaystyle\int \left( \frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x \right) \, \mathrm{d}x \\&= \frac {x^8}{8} - \frac {2}{3} x^6 + 4x^4 - 64x + 256 \displaystyle\int \frac {x}{x^2 + 4} \, \mathrm{d}x. \end {align*} $$The last integral can be done with an easy substitution.


Edit: So apparently the problem got changed again, but the same idea applies. You can re-write the new integrand as $$ \frac {x^9}{(x^2+4)^6} = \frac {x}{(x^2+4)^2} - \frac {16x}{(x^2+4)^3} + \frac {96x}{(x^2+4)^4} - \frac {256x}{(x^2+4)^5} + \frac {256x}{(x^2+4)^6}, $$ and use the substitution $u=x^2+4$ on each of these.

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  • $\begingroup$ sorry see the edit $\endgroup$ – RE60K Aug 22 '14 at 19:16
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    $\begingroup$ @Aditya You can still use the same approach. Edited. $\endgroup$ – Ahaan S. Rungta Aug 22 '14 at 19:19
  • $\begingroup$ Now this is long, try to match with given answers[see my edit] $\endgroup$ – RE60K Aug 22 '14 at 19:29
  • $\begingroup$ @Aditya Hm, Tunk-Fey's answer seems to be better suited now that you added the multiple-choice. You should accept his answer! ;) $\endgroup$ – Ahaan S. Rungta Aug 22 '14 at 19:39
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$$I=\int \frac{x^9}{(x^2+4)^6}dx=\frac{1}{2}\int \frac{2x(x^2)^4}{(x^2+4)^6}dx=$$ $$x^2+4=u,x^2=u-4,2xdx=du$$ $$=\frac{1}{2}\int \frac{(u-4)^4}{u^6}du$$

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    $\begingroup$ The problem got changed. See my edited answer. $\endgroup$ – Ahaan S. Rungta Aug 22 '14 at 19:22
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    $\begingroup$ After the edit, this answer becomes OP's first substitute :) $\endgroup$ – Kaster Aug 22 '14 at 19:23
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The substitution $t=1+ 4/x^2$ seems to be useful in your case.

We have $$x^2+4 = \frac{4t}{t-1},$$ $$x^9 \frac{dx}{dt} = - \frac18 x^{12} = -\frac18\frac{4^6}{(1-t)^6},$$ and thus $$\int \frac{x^9}{(x^2+4)^6} dx = - \frac18 \int t^{-6} dt. $$

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  • $\begingroup$ ^simply because, if you see all the options, you see one thing common. That is $1+4/x^2$. So this is obvious substitution(alteast you have lot of motivation to perform this substitution). Had it be a subjective problem, thinking this substitution would be much harder $\endgroup$ – Shivang jindal Aug 23 '14 at 13:51

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