Help in finding the integral function.

Question

Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.

Partial fractions should work (though it looks unpleasant). $x^3+1$ factors. — lulu, Jul 31 '15 at 19:42
Rewriting the $x^3 + 1$ in the denominator as $(x+1)(x^2-x+1)$ (which I found via long division) should help out! — Khallil, Jul 31 '15 at 19:46
I'd suggest you not to do this by hand. Did someone give you this as an exercise or are you just curious? — mickep, Jul 31 '15 at 19:50
Mathematica gives the following partial fraction decomposition, which looks horrible as expected: $$ \scriptsize \frac{1}{(x^3+1)^3} = \frac{8-7 x}{27 (x^2-x+1)^2}+\frac{7-5 x}{27(x^2-x+1)}+\frac{1-2 x}{9 (x^2-x+1)^3}+\frac{5}{27 (x+1)}+\frac{1}{9 (x+1)^2}+\frac{1}{27 (x+1)^3}. $$ So... good luck! — Sangchul Lee, Jul 31 '15 at 19:52
@SangchulLee , what's ugly about that? It looks nice! Let's not get dumbed down by Mathematica and Wolfram Alpha! — Count Iblis, Jul 31 '15 at 19:53
Integration by parts gives you a recurrence relation for $\int \frac{1}{(x^3+1)^k},dx$. The case $k = 1$ isn't too bad, and the recurrence is probably a bit less work than the direct partial fraction decomposition. — Daniel Fischer, Jul 31 '15 at 19:55

score 1 · Answer 1 · answered Jul 31 '15 at 20:05

1

Remember that $x^3+1 = (x+1)(x^2-x+1)$. That quadratic polynomial cannot be factored using real numbers, since the discriminant ($b^2-4ac\vphantom{\dfrac\int\int}$) is negative. So $(x^3+1)^3 = (x+1)^3 (x^2-x+1)^3$. That should tell you what the partial fraction decomposition is.

To handle $x^2-x+1$, one (of course) completes the square: \begin{align} x^2-x+1 & = \left( x^2 - x + \frac 1 4\right) + \frac 3 4 \\[10pt] & = \left( x - \frac 1 2 \right)^2 + \frac 3 4 \\[10pt] & = \frac 3 4 \left( 1 + \left( \frac{2x-1}{\sqrt 3} \right)^2 \right). \end{align}

Then $\tan\theta = \dfrac{2x-1}{\sqrt 3}$ and $1+\tan^2\theta=\sec^2\theta$ and $\sec^2\theta\,d\theta = \dfrac{2\,dx}{\sqrt 3}$, etc.

answered Jul 31 '15 at 20:05

Michael Hardy

1

How would opting for complex factors fare in integration? Would we need to make any extra considerations? – Khallil Jul 31 '15 at 20:14
@Khallil : As far as algebra and arithmetic go, no problem. But if you evaluate $\displaystyle\int\frac {dx} {x-c}$ where the complex number $c$ is not real, you have to contend with things like the multiple-valued nature of the complex logarithm function, i.e. $z = e^w = e^{w+2\pi i n}$, so $\log z = w +2\pi in$ for $n=0,\pm1,\pm2,\ldots$. In a problem in which this logarithm occurs repeately, this gets rather involved. One works with definite integrals not just going from $a$ to $b$, i.e. $\displaystyle\int_a^b$, but${},\ldots\qquad{}$ – Michael Hardy Jul 31 '15 at 20:31
$\ldots,{}$ integrals along a path that may start at some point $a$ and wind some number of times around $c$ (the number that appeared in $1/(x-c)$) and the number of times the path winds around $c$ matters. ${}\qquad{}$ – Michael Hardy Jul 31 '15 at 20:35
Thanks for the answer, @MichaelHardy. I'm guessing this is something that one would encounter in a complex analysis course? (Also, would the number of times (say $n$) the path winds around $c$ mean that the final result would be $n$ lots of the integral considered?) – Khallil Jul 31 '15 at 22:59
This is covered in complex analysis courses. With $\displaystyle\int\frac{dz} z$, integrating over a path that starts at $1$ and ends at $x\in\mathbb C$ gives one of the values of the multiple-valued complex logarithm, and which one it is depends on how many times you wind around $0$. In $\displaystyle\int \frac{f(z)}{z-c},dz$, if you follow the same path every time you wind around $c$, you do add the same amount every time, and if $f(z)$ is everywhere differentiable (including at $c$) then it doesn't even matter if you follow the same path every time. ${}\qquad{}$ – Michael Hardy Aug 01 '15 at 00:01

score 1 · Accepted Answer · answered Jul 31 '15 at 20:22

This following trick, which is essentially the same as @Daniel Fischer's suggestion, reduces your burden of calculating partial fraction decomposition. First, write

$$ \int \frac{dx}{(1+x^3)^3} = \int \frac{1 + x^3}{(1+x^3)^3} \, dx - \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx. $$

Then by integration by parts, we have

$$ \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx = - \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{1}{6}\int \frac{dx}{(1+x^3)^2}. $$

Plugging this back to the first identity, we get

$$ \int \frac{dx}{(1+x^3)^3} = \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{5}{6} \int \frac{dx}{(1+x^3)^2}. $$

Similarly, we have

$$ \int \frac{dx}{(1+x^3)^2} = \frac{1}{3} \frac{x}{(1+x^3)^2} + \frac{2}{3} \int \frac{dx}{1+x^3}. $$

Finally, the last integral can be calculated easily by partial fraction decomposition.

Help in finding the integral function.

2 Answers2