Find, without partial fractions $$\int\dfrac{1}{x^3+1}dx$$

My Attempt: I was able to do it via partial fractions by factoring the denominator as


However, I then tried a different approach without using partial fractions. I added and subtracted $+x^3$ in the numerator and wrote the integrand as


Then, as the first term is easily integrable, I took the second term and wrote it as

$$\dfrac{x^2\cdot x}{x^3+1}.$$

Using Integration by Parts, I integrated


and differentiated $x$. I ended up with a term and a new integral,

$$\dfrac{x\cdot \ln{(x^3+1)}}{3} + \int \dfrac{\ln{(x^3+1)}}{3}dx$$

To evaluate the second integral, I again used integration by parts wherein I integrated $x$ and differentiated


Finally, I got the original integral as one of the parts. However, when I undid all the integration by parts to substitute in the original integral, both sides had the same terms and I ended up with

$$0 = 0.$$

Is there any other way to solve this integral?

  • 2
    $\begingroup$ See here for a tutorial about how to use MathJax. $\endgroup$
    – DMcMor
    Jul 18, 2019 at 17:18
  • $\begingroup$ the partial fractions approach here is certainly the easiest $\endgroup$
    – fGDu94
    Jul 18, 2019 at 17:21
  • 1
    $\begingroup$ Do you mean you did something like this? $\int f'g=fg-\int g'f=fg-\left(fg-\int f'g\right)$ $\endgroup$
    – user239203
    Jul 18, 2019 at 17:21
  • $\begingroup$ @Gae.S. yeah I used by parts twice $\endgroup$
    – user671231
    Jul 18, 2019 at 17:25
  • $\begingroup$ @GeorgeDewhirst yes I know that. I just want to learn and explore other methods to expand my arsenal or integration skills $\endgroup$
    – user671231
    Jul 18, 2019 at 17:26

1 Answer 1


Yes, there is a way without partial fraction. This might also be in your interest.

Start off with the substitution: $$x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ This substitution produces a nice cancelation in the denominator, since: $$(1+t)^3+(1-t)^3=1+3t +3t^2+t^3 +1-3t+3t^2 -t^3=2(1+3t^2)$$

$$\int \frac{1}{1+x^3}dx=-\int\frac{1}{\frac{(1+t)^3}{(1+t)^3}+\frac{(1-t)^3}{(1+t)^3}}\frac{2}{(1+t)^2}dt=-\int \frac{1+t}{1+3t^2}dt$$ $$=-\frac{1}{\sqrt 3}\arctan(\sqrt 3 t)-\frac16 \ln(1+3t^2)+C,\quad t=\frac{1-x}{1+x}$$

  • 1
    $\begingroup$ Thanks a lot for the answer $\endgroup$
    – user671231
    Jul 19, 2019 at 13:51

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