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 at 17:18
  • $\begingroup$ the partial fractions approach here is certainly the easiest $\endgroup$ – George Dewhirst Jul 18 at 17:21
  • 1
    $\begingroup$ Do you mean you did something like this? $\int f'g=fg-\int g'f=fg-\left(fg-\int f'g\right)$ $\endgroup$ – Gae. S. Jul 18 at 17:21
  • $\begingroup$ @Gae.S. yeah I used by parts twice $\endgroup$ – StackUpPhysics Jul 18 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 $\endgroup$ – StackUpPhysics Jul 18 at 17:26

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 $\endgroup$ – StackUpPhysics Jul 19 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.