Evaluate $\int\frac{dx}{x^3+1}$ without using partial Fractions Evaluate $$I=\int\frac{dx}{x^3+1}$$ without using partial Fractions
I tried in this way: 
$$I=\int \frac{\frac{xdx}{x^4}}{1+\frac{1}{x^3}}$$
Put $1+\frac{1}{x^3}=t$
$$I=\frac{-1}{3}\int \frac{dt}{t(t-1)^{\frac{1}{3}}}$$
Any way to help me?
 A: hint
Write the integrand as
$$\frac {1}{x^3+1}=1-\frac {x}{3}\frac{3x^2}{x^3+1}$$
and integrate by parts.
A: If you really want to do it, you can do the following. This is really just a sketch, there are lots of fine details omitted.
Write 
$$\int \frac{1}{x^3+1} dx = \int 1 - \frac{x^3}{x^3+1} dx = x - \frac{1}{3} \int x \frac{3x^2}{x^3+1} dx.$$ 
This second term can be integrated by parts: take $u=x,dv=\frac{3x^2}{x^3+1}$. The boundary term gives a contribution to the integral and leaves you to evaluate $\int \ln(x^3+1) dx$. This can be split into $\ln(x+1) + \ln(x^2-x+1)$. The first term is easy to integrate by parts. The second term is $\ln((x-1/2)^2+3/4)$. Upon a suitable change of variable and factoring out constants, it becomes enough to compute $\int \ln(x^2+1) dx$. This can be done by integrating by parts the opposite of how we did before: $dv=dx,u=\ln(x^2+1)$ forces us to evaluate $\int x \frac{2x}{x^2+1} dx$, but this is now easy after a long division. 
This is way harder than doing it by partial fractions, basically because you wound up doing partial fractions implicitly anyway, but obfuscated by two integration by parts steps.
