0
$\begingroup$

This is the indefinite integral I have to evaluate: $$\int\frac{x^3}{x^2-x-2}dx$$ so by using the long division on polynomials technique, I got to: $$\frac{x^2}{2}+x+\int\frac{3x+2}{x^2-x-2}dx$$ How do I continue from here? I thought of using "integration of rational functions" but it didn't work.

$\endgroup$
  • 1
    $\begingroup$ Factor the denominator and use partial fractions. $\endgroup$ – user31415926535 Jul 2 '15 at 11:52
  • 5
    $\begingroup$ Partial fractions, write ${3x+2\over x^2-x-2}={A\over x-2}+{B\over x+1}$. $\endgroup$ – David Mitra Jul 2 '15 at 11:52
  • $\begingroup$ What didn't work in "integration of rational functions" ? This IS a rational function. $\endgroup$ – Yves Daoust Jul 2 '15 at 12:09
  • $\begingroup$ @DavidMitra thaks! (I solved) $\endgroup$ – Yagel Jul 2 '15 at 12:12
4
$\begingroup$

Specifically, you have that $$\frac{3x+2}{x^2 - x - 2} = \frac{A}{x-2} + \frac{B}{x+1}$$

The usual technique reveals that $A = \frac{8}{3}$ and $B = \frac{1}{3}$


Hence $$ \begin{align}\frac{x^3}{x^2 - x - 2} &= x + 1 + {\color{blue}{\frac{3x+2}{(x-2)(x+1)}}} \\ \\ &= x + 1 + \frac{8}{3(x-2)} + \frac{1}{3(x+1)} \end{align}$$

So that $$\begin{align} \int \frac{x^3}{x^2 - x - 2} \, \mathrm{d}x &= \int x + 1 + {\color{blue}{\frac{3x+2}{x^2 - x - 2}}} \\ &= \int x + 1 + \frac{8}{3(x-2)} + \frac{1}{3(x+1)} \, \mathrm{d}x \end{align}$$

Hence $$\int \frac{x^3}{x^2 - x - 2} = x + \frac{x^2}{2} + \frac{8}{3}\ln |x-2| + \frac{1}{3} \ln |x+1| + \mathrm{C}$$

$\endgroup$
0
$\begingroup$

Hint : $\frac{3x+2}{x^2-x-2}=\frac{3}2\frac{2x-1}{x^2-x-2}+\frac{7}2\frac{1}{x^2-x-2}$ (other useful forms are possible as pointed out in the comments)

$\endgroup$

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.