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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$\begingroup$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:
$\displaystyle \int\frac{1}{(ax+b)^n}\,dx$, with $a,b$ constants, $a\neq 0$, and $n\geq 1$.
$\displaystyle \int\frac{x}{(ax^2+bx+c)^n}\,dx$, with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.
$\displaystyle \int\frac{1}{(ax^2+bx+c)^n}\,dx$ 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.
$\endgroup$ $\begingroup$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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