14
$\begingroup$

I saw this method in some random PDF and am intrigued of the exact method used. I can't find any page of this method on the web because I'm not sure what you'd call this method.

Here is the method:

For solving $$\int \frac{2x^4 + x^3}{x^2 + x - 2} \,\text{d}x$$ Observe that $$\begin{align*} 2x^4 + x^3 &= 2x^2 (x^2+x-2) - x^3+4x^2 \\ &= 2x^2 (x^2+x-2) - x(x^2+x-2) + 5x^2-2x \\ &= 2x^2 (x^2+x-2) - x(x^2+x-2) + 5(x^2+x-2) - 7x+10 \\ &= (2x^2-x+5)(x^2+x-2) - 7x+10 \end{align*}$$ and then $$ \int \frac{2x^4-x^3}{x^2+x-2} \,\text{d}x = \int (2x^2-x+5)\,\text{d}x + \int \frac{-7x+10}{x^2+x-2}\,\text{d}x$$

$\endgroup$
7
  • 17
    $\begingroup$ In fact it is precisely the (Euclidean) polynomial division algorithm, except it's performed explicitly in equational form, rather than having the equations be implicit in some tabular format. $\endgroup$
    – Bill Dubuque
    Jan 15, 2017 at 23:52
  • 2
    $\begingroup$ @TheGreatDuck No, that requires you to factor the denominator, which was never done. $\endgroup$
    – Simply Beautiful Art
    Jan 16, 2017 at 0:56
  • $\begingroup$ @BillDubuque I had seen the implicit algorithm before, but it is much more enlightening to see it done this way. Are there any sources you could reccommend that explore this topic further? $\endgroup$
    – Ovi
    Mar 8, 2017 at 18:02
  • $\begingroup$ @Ovi What precisely are you interested in exploring further? $\endgroup$
    – Bill Dubuque
    Mar 8, 2017 at 18:42
  • $\begingroup$ @BillDubuque Sorry for the ambiguity, I am wondering if you have any nice sources about how various implicit algorithms have been derived, and which is the most efficient one. $\endgroup$
    – Ovi
    Mar 8, 2017 at 18:49

1 Answer 1

Reset to default
40
$\begingroup$

What you have is literally polynomial long division written out without division signs. Indeed, what is written out here is the essence of polynomial long division, which is all about finding the coefficient of the factor that returns the highest degree term in the original polynomial.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .