Source: A question bank on challenging integral problems for high school students.

Problem: Evaluate the indefinite integral $$\int{dx\over{(x^3+1)^3}}$$

Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fractions but it would be VERY time consuming and altogether boring. I am unable to find an alternate solution other than partials. Can anyone lead me towards a quicker approach, because the exam I'm preparing for is time bound and I can't afford to spend much time on a single problem. Thanks!

Edit 1:

I looked upon reduction formulas. I guess we can generalise this by using:

$$I_n = \int {dx\over{(x^3+1)^n}}$$

$$I_n = \int{(x^3+1)^{-n}dx}$$

Will try and solve it. Maybe we can reduce it to a simpler integral!

Edit 2:

Ok so I got the reduction formula. Can anybody verify if its right, like if you've solved it?

$$I_{n+1} = \frac {x}{3n(x^3+1)^n} + \frac{(3n-1)}{3n}I_n$$

Edit 3: Now I have reached the solution nearly

take $n=2$

$$I_3 = \frac{x}{6(x^3+1)^2}+\frac{5}{6}I_2$$

now take $n=1$

$$I_2 = \frac{x}{3(x^3+1)}+\frac{2}{3}I_1 $$

now we see that $I_1$ is nothing but

$$I_1= \int{\frac{dx}{x^3+1}}$$

which simplifies to

$$I_1 = \int{\frac{dx}{(x+1)(x^2+1-x)}}$$

Now this integral is cake, solve using Partial Fractions

$$\frac{1}{(x+1)(x^2+1-x)} \equiv \frac{A}{x+1} + \frac{Bx+C}{(x^2+1-x)}$$

And eventually $I_1$ is:


This way we get $I_1$. Putting in equation above we get $I_2$

Substitute $I_2$ in the equation above and obtain an expression for $I_3$ !

I'll verify ASAP :)

Final edit: Yeas! Reached the answer. Matches term to term! Final answer is :


Use the above approach or any alternatives that are more quick are welcome!

  • $\begingroup$ It seems that partial fractions might be the only method of solution, here. C.f. m.wolframalpha.com/input/… $\endgroup$ Aug 22 '17 at 15:42
  • $\begingroup$ That's a pretty brutal problem for high school students. $\endgroup$ Aug 22 '17 at 15:46
  • 1
    $\begingroup$ You can save some steps, by taking $a=1$ in $$ \frac{1}{2}\frac{d^2}{da^2}\int \frac{dx}{a+x^3}$$. $\endgroup$
    – Paul Enta
    Aug 22 '17 at 15:50
  • 1
    $\begingroup$ @YourAverageEuler Can you use all complex numbers? If not, then partial fractions won't work for finding $\int \frac{dx}{\left(x^2-x+1\right)^3}$. Notice that $x^2-x+1$ has no real roots. $ax^2+bx+c$ with $b^2-4ac<0$ has no real roots. WolframAlpha (link) agrees that you need imaginary numbers here. This link (link) has a method for finding $\int \frac{dx}{\left(x^2-x+1\right)^3}$. I could show an easier method. $\endgroup$
    – user236182
    Aug 23 '17 at 13:18
  • 1
    $\begingroup$ @YourAverageEuler In the last link I gave (link), see the example $\int \frac{1}{\left(x^2+1\right)^2}\, dx$. It generalizes. Notice that $$\left(\frac{1}{\left(x^2+bx+c\right)^n}\right)'=\frac{-(2x+b)n}{\left(x^2+bx+c\right)^{n+1}}$$ $\endgroup$
    – user236182
    Aug 23 '17 at 13:22

We have:

$$\int\frac{dx}{(x^3 + 1)^3} = \int \frac1{x^2}\frac{x^2}{(x^3 + 1)^3}dx$$


$$u = \frac1{x^2}, dv = \frac{x^2}{(x^3 + 1)^3}dx$$

this becomes:

$$\text{something } - \frac13\int\frac{1}{x^3(x^3 + 1)^2}dx$$


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

do same trick for $\int \frac{dx}{(x^3 + 1)^2}$ to get:

$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \text{something}' + \frac53 \int\frac{dx}{x^3(x^3 + 1)}$$

so this reduces to finding

$$\int\frac{dx}{x^3 + 1} $$

this is:

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

integrate by parts and voila!

  • $\begingroup$ thanks @cauchy, i will solve it and upload the answer asap $\endgroup$ Aug 23 '17 at 2:21
  • $\begingroup$ Can you please breakdown the second step of your solution? I'm pretty much a noob :p $\endgroup$ Aug 23 '17 at 2:31
  • $\begingroup$ @YourAverageEuler sorry, I'm pretty tired :( Maybe I'll do that later $\endgroup$
    – Cauchy
    Aug 23 '17 at 2:33
  • $\begingroup$ How about we create a reduction formula?! $\endgroup$ Aug 23 '17 at 6:43

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