Real VS Complex for integrals: $\int_0^\infty \frac{dx}{1 + x^3}$ The integral
$$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$
can be evaluated both by a complex method (residues) and by a
real method (partial fraction decomposition).
The complex method works also for the integral 
$$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$
but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral.

Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\  \text{as} \ N \to \infty$$ while
$$\eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \  \text{as} \ N \to \infty\cr}$$
and (if $b > 0$)
$$ \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1)
 \  \text{as} \ N \to \infty\cr}$$
In particular, from the partial fraction decomposition
$$ \frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} +
\frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$
you get
$$ \int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3}  + o(1)$$
i.e.
$$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
