So I want to find all antiderivaties of


Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.

I've started by doing:

$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$


$x = A(x-1)+B(x^2+x+1)$

Then setting $x = 1$ to get $3B = 1 => B = \frac{1}{3}$

So I have $\int({\frac{A}{x^2+x+1}}+{\frac{1}{3(x-1)}})$

However, I'm not sure how to continue from here.

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    $\begingroup$ Your $A$ in the first step should be an $Ax+C$ since the polynomial in the denominator is a quadratic. $\endgroup$ – Joe Johnson 126 Jan 10 '11 at 15:34
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    $\begingroup$ Complete the square in the denominator of the first term and then try a trig substitution. $\endgroup$ – Qiaochu Yuan Jan 10 '11 at 15:38
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    $\begingroup$ @Moron: Yes; but it's important to note that trying it with only constants in the numerator may lead to impossibilities in the general case. I don't know if Mikael did not use a linear numerator because he already knew he did not need one, or because he is making a standard error of thinking that you only put "unknown constants" in the numerators of a partial fraction decomposition. $\endgroup$ – Arturo Magidin Jan 10 '11 at 17:24
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    $\begingroup$ @Moron: I'm sorry, but now I'm utterly confused. If you have, say, $\frac{ax^2+bx+c}{(x-r)(x^2+sx+t)}$, with $x^2+sx+t$ irreducible quadratic, then one should, in my opinion, do $$\frac{ax^2+bx+c}{(x-r)(x^2+sx+t)} = \frac{A}{x-r} + \frac{Bx+C}{x^2+sx+t}.$$ The OP dropped the corresponding $Bx$. In some cases, one may be able to get away with trying $\frac{A}{x-r}+\frac{C}{x^2+sx+t}$ and get a solution, but usually one will not get a solution. Some people are saying he should use the correct form regardless as good practice. You seemed to be saying there is no point... $\endgroup$ – Arturo Magidin Jan 10 '11 at 17:43
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    $\begingroup$ @Moron: Yes, for this problem there is no point. But we don't know if the OP did not do it because he knew there was no point in this problem, or if he did not do it because he is committing a very common error of always using constants in the numerator, even with irreducible quadratic numerators. $\endgroup$ – Arturo Magidin Jan 10 '11 at 17:52

You can do partial fractions in this way: $$\frac{x}{x^{3}-1} = \frac{Ax+B}{x^{2}+x+1} + \frac{C}{x-1}$$

Now by equating the coefficients you get:

  • $A+C=0$

  • $B+C-A=1$ and;

  • $C-B=0$.

Solving you get the values of $A= -\frac{1}{3}$, $B=\frac{1}{3}$ and $C=\frac{1}{3}$.

So, the required integral breaks down as $$ \int\frac{x}{x^{3}-1} \ dx = -\frac{1}{3}\int \frac{x -1}{x^{2}+x+1} \ dx + \frac{1}{3} \int \frac{1}{x-1} \ dx$$

Now $$\int\frac{x-1}{x^{2}+x+1} \ dx = \frac{1}{2} \int \frac{2x-2}{x^{2}+x+1} \ dx =\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1} -\frac{1}{2} \int \frac{3}{x^{2}+x+1} \ dx$$

The first part of the integral is easy. For the second part use $$x^{2}+x+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \Bigl(\frac{\sqrt{3}}{2}\Bigr)^{2}$$ and then put $t=x+\frac{1}{2}$ and then the trignometric substitution.

  • $\begingroup$ Thanks for your answers all of you. However, I fail to see how $\frac{1}{2} \int \frac{2x-2}{x^{2}+x+1} \ dx =\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1} + \frac{1}{2} \int \frac{3}{x^{2}+x+1} \ dx$ computes, wrong sign before the last integral to the right, no? $\endgroup$ – JaHei Jan 10 '11 at 17:51
  • $\begingroup$ @Mikael E: Sorry mikael it shoud be -3 on the second integral. $\endgroup$ – anonymous Jan 10 '11 at 17:59

Like most techniques of integration, the ideal of partial fractions is to reduce a difficult integral to integrals that are, if not easy, at least doable. The three types of integrals that show up when doing partial fractions are:

  1. $\displaystyle \int\frac{1}{(ax+b)^n}\,dx$, with $a,b$ constants, $a\neq 0$, and $n\geq 1$.

  2. $\displaystyle \int\frac{x}{(ax^2+bx+c)^n}\,dx$, with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.

  3. $\displaystyle \int\frac{1}{(ax^2+bx+c)^n}\,dx$ with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.

The first type is easy to solve: do a substitution $u=ax+b$ and go at it.

Second type is a bit more involved. By doing a substitution $u=ax^2+bx+c$, we get \begin{align*} \int\frac{x}{(ax^2+bx+c)^n}\,dx &= \frac{1}{2a}\int \frac{2ax\,dx}{(ax^2+bx+c)^n}\\ &= \frac{1}{2a}\left(\int\frac{2ax+b}{(ax^2+bx+c)^n}\,dx - \int\frac{b}{(ax^2+bx+c)^n}\,dx\right)\\ &= \frac{1}{2a}\int\frac{du}{u^n} - \frac{b}{2a}\int\frac{dx}{(ax^2+bx+c)^n}. \end{align*} The first integral can be done; the second reduces to the third type mentioned above.

And so we come down to the third type. When $n=1$, the simplest thing to do is to complete the square; factoring out $a$ we may assume we have $x^2+Bx+C$. Completing the square, you get $(x+\frac{B}{2})^2 + (C - \frac{B^2}{4})$. Because we are assuming that the original quadratic is irreducible, that means that $B^2 - 4C\lt 0$, so that $C-\frac{B^2}{4}\gt 0$. Substituting $u=X+\frac{B}{2}$ turns this into a fraction of the form $\frac{1}{u^2+r^2}$; factor out $r^2$, do another substitution, and you can turn it into a fraction of the form $\frac{1}{w^2+1}$. But $\int\frac{dw}{w^2+1}$ is an easy integral: you have an immediate antiderivative for it.

So, modulo a bunch of algebra and some substitution, you can solve $\int\frac{1}{ax^2+bx+c}\,dx$ with $ax^2+bx+c$ irreducible quadratic: complete the square, do some substituttions, and turn it into $\int\frac{1}{w^2+1}\,dx$.

What if you have $\int\frac{dx}{(ax^2+bx+c)^n}$ with $n\gt 1$, $ax^2+bx+c$ irreducible quadratic? Those are more complicated, but not too bad; you can still complete the square and do a bit of algebra, so that you bring it to the form $$\int \frac{du}{(u^2+r^2)^n}$$ for some positive $r$. Then one can use the reduction formula (obtained by doing integration by parts): $$\int\frac{du}{(u^2+r^2)^n} = \frac{1}{2r^2(n-1)}\left(\frac{u}{(u^2+r^2)^{n-1}} + (2n-3)\int\frac{du}{(u^2+r^2)^{n-1}}\right)$$ and continuing this way you will eventually end in the integral with denominator $u^2+r^2$, which we know how to do.


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