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When evaluating the integral $\int \frac{x^2 + 8}{x^3 + 9x} \textrm{d}x$, WolframAlpha gave an interesting step-by-step suggestion. Here is the suggested method:

$$\int \frac{x^2 + 8}{x^3 + 9x} \textrm{d}x = \int \frac{x^{17} + 8x^{15}}{x^{18} + 9x^{16}} \textrm{d}x=\frac{1}{18} \int \frac{18(x^{17} + 8x^{15})}{x^{18} + 9x^{16}} \textrm{d}x$$

Let $u = x^{18}+9x^{16} \implies \textrm{d}u=(18x^{17}+144x^{15}) \, \textrm{d}x = 18(x^{17} + 8x^{15}) \, \textrm{d}x $

$$\frac{1}{18} \int \frac{\textrm{d}u}{u} = \frac{1}{18} \log \vert u \vert + C $$

This indeed works, giving the correct answer (after substituting $x$ back and simplifying). But I've never seen this kind of an integration technique, and I'm wondering if there's a name for this method and perhaps a systematic way of determining which degree of polynomial to use? (The polynomial in this example happened to be $x^{15}$, something that I would've never though of)

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I'm not sure what this method is called, but we can explore how to use it in general.

Consider integrals of the general form, $$ \int \frac{a x^{n+k}+bx^{n}}{cx^{n+k+1}+dx^{n+1}}\textrm{d}x $$

We will pick $u = x^{p}\cdot(cx^{n+k+1} +dx^{n+1})$ and want $\textrm{d}u = C x^p\cdot(ax^{n+k}+bx^n)$ for some constants $C$ and $p$.

Note that we have $$ \textrm{d}u = c\cdot(n+k+p+1) x^{n+k+p} + d\cdot(n+p+1) x^{n+p} $$

So we want to choose $C$ and $p$ such that, $$ C\cdot a = c\cdot(n+k+p+1), ~~~C\cdot b = d\cdot(n+p+1) $$

In your example, $n=0$, $k=2$, $a=c=1$, $b=8$, and $d=9$ so we would solve, $$ C = (0+2+p+1) = p+3, ~~~ C\cdot 8 = 9\cdot(0+p+1) = 9p+9 $$

This is a linear system in two variables which you can solve to find $p=15$ and $C=18$.

Note that we could have the denominator have a degree more than one above the numerator and do a similar process to solve.

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  • $\begingroup$ Quick note, I think you're missing an $x$ from the $d$ term in $u=x^p \cdot (c x^{n+k+1}+d^{n+1})$. But thank you, very interesting answer. And thanks especially for taking the time to do it in general. $\endgroup$ – imas145 Jan 7 at 17:04

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