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
$$(x+1)(x^2-x+1)$$
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
$$1-\dfrac{x^3}{x^3+1}.$$
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
$$\dfrac{x^2}{x^3+1}$$
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
$$\ln{(x^3+1)}.$$
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?