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)


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.

| cite | improve this answer | |
  • $\begingroup$ 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 '19 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.