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So I want to find all antiderivaties of

$\frac{x}{x^3-1}$

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:

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

Hence,

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

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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4  
Your $A$ in the first step should be an $Ax+C$ since the polynomial in the denominator is a quadratic. –  Joe Johnson 126 Jan 10 '11 at 15:34
    
After the fix you can expand the second equation and equate coefficients. –  Joe Johnson 126 Jan 10 '11 at 15:35
2  
Complete the square in the denominator of the first term and then try a trig substitution. –  Qiaochu Yuan Jan 10 '11 at 15:38
    
@Joe: Unless you get $A=2B$, setting the numerator to $Ax+B$ does not really help, you have to compute the integral of $\frac{1}{1+x+x^2}$ anyway... –  Aryabhata Jan 10 '11 at 17:14
2  
@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. –  Arturo Magidin Jan 10 '11 at 17:24

2 Answers 2

up vote 6 down vote accepted

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.

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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? –  Mikael E Jan 10 '11 at 17:51
    
@Mikael E: Sorry mikael it shoud be -3 on the second integral. –  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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