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Evaluate $$\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}\mathrm dx$$

In the above question, I was literally stumped, and wasn't able to solve it for a long time. Turns out that you had to divide the numerator and the denominator by $x^{15}$, and then we could substitute. Now this got my thinking, how can we understand where to divide what? I mean obviously, in this question - one big hint would be the numerical coefficients of the numerator, but other than that is there any logical way to proceed or is it basically a hit and try?

[Link to the solution] 1

Integreek
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AnonMouse
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3 Answers3

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Generally these type of questions we simply put $\displaystyle x = \frac{1}{t}$ and then use normal substution method

Now let $$I = \int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx$$

Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$

So $$I = -\int\frac{2+5t^3}{\left(t^5+t^2+1\right)^3}\cdot \frac{t^{15}}{t^{12}}\cdot \frac{1}{t^2}dt = -\int\frac{2t+5t^4}{\left(1+t^2+t^{5}\right)^3}dt$$

Now using normal substution method, Put $(1+t^2+t^5)=u\;,$ Then $(2t+5t^4)dt=du$

So we get $$I = -\int\frac{1}{u^3}du = \frac{1}{2u^2}+\mathcal{C} = \frac{1}{2(1+t^2+t^5)^2}+\mathcal{C}$$

So $$I = \int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx = \frac{x^{10}}{2(x^5+x^3+1)^2}+\mathcal{C}$$

$\bf{Bonus\; Question}::$ Evaluation of $$\int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx$$

juantheron
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  • Thanks a lot! It is certainly an interesting way to solve this question. I am attempting the bonus question, but can you please elaborate the reasoning behind doing this substitution. Also, where all does this substitution work. – AnonMouse Jul 12 '16 at 13:38
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    Got it! The answer for the bonus question should be $ \frac{x^5}{x^5 + x + 1} + C $. – AnonMouse Jul 12 '16 at 13:53
  • http://math.stackexchange.com/questions/1395564/find-int-frac5x44x5x5x12/1395569#1395569 – juantheron Jul 12 '16 at 14:08
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Such problems are designed where author of the problem has the tool to be applied in mind and the problem is tailor-made to force the application of the tool. There are much more intense pathological problems devised to make the application of that tool all the more obscure. I remember problems by Titu Andreescu where he wants the student to apply simple AM-GM inequality which is so obscurely hidden behind many steps of algebraic manipulation, that logical approach seems difficult. In your case, its relatively easy to see that leading term of denominator is $x^{15}$, so divide by it as a rule of thumb (you want to deal with lower powers) and then use the second rule of thumb when integration is not straightforward--change of variables.

user_1_1_1
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Usually, a method that can work(which fortunately does work in this case) to evaluate integrals of rational functions where the denominator is a polynomial raised to some power is multiplying and dividing by some suitable power of $x$ such that the derivative of the aforesaid polynomial appears in the numerator.

This is usually the highest power of $x$ that can be taken common from the denominator, i.e., $15$ in this case. Say this didn’t strike you, then if the required power of $x$ is $3p$, we desire for some $k\in\mathbb R$:

$$(p+5)x^{p+4}+(p+3)x^{p+2}+px^{p-1}=k\left(2x^{3p+12}+5x^{3p+9}+0\right)$$

Now here, we have $3$ choices corresponding to the term $0$ on the RHS: $p=-5, p=-3, p=0$. It can be easily seen that only for $p=-5$, $p+3:p=2:5\implies k=-1$ which is why we should divide by $x^{15}$ in the numerator and the denominator.

P.S. This method doesn't work in this integral as we have to multiply and divide by $-x^2+3x-\frac{26}9$ amd not simply a power of $x$, which is why I said this method may or may not work for such integrals. But for those integrals where it works, it simplifies the integration beautifully.

Integreek
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