Finding the integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}dx$ with substitution - how to think?

Question

Find

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

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 x¹⁵, 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

Answer 1

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 $$\int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx$$

Answer 2

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.