Finding $\int \frac{1}{1+x^3}dx$ without partial fractions

Question

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?

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: $$(1+t)^3+(1-t)^3=1+3t +3t^2+t^3 +1-3t+3t^2 -t^3=2(1+3t^2)$$

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