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$$int_{0}^{infty} 1/(x^3+1) dx$$ the question says use complex functions to answer this. I tried using Cauchy residue theorem and this $$1/2int_{-infty}^{infty} 1/(x^3+1) dx=int_{0}^{infty} 1/(x^3+1) dx$$
So far i got that the poles are π/3,π.. but only π/3 matters as is it is inside the contour of integration. I then used cauchy residue theorem and did not get the answer of 2π/(3sqrt3). Is this the method i should be using? or is there an alternative way using complex functions. The intuition i used for this exercise was by looking at an example with $$int_{0}^{infty} 1/(x^4+1) dx$$.

You made an error here:

$$ frac12 int^infty_infty frac1{1 + x^3} stackrel{??}{=} int_0^infty frac1{1 + x^3}$$

This is not valid in this case because $frac1{1 + x^3} neq frac1{1 + (-x)^3}$ in general (this is what was used for $1/(1 + x^4)$ - there it works because $x^4$ is even).

The trick I know for doing this integral is to integrate along the border of a circle segment with angle $2 pi/3$ (see Frank's answer).

Let's generalize this by denoting$$mathfrak{I}(n)=intlimits_0^{infty}frac {mathrm dx}{1+x^n}$$where $ngeq2$ and let$$f(z)=frac 1{1+z^n}$$The trick for $mathfrak{I}(n)$ is to integrate it along a sector instead of a circle or half circle. Let the radius of the sector be $R$ and the arc integral to be denoted as $Gamma_R$. Since we only want to enclose one singularity, we'll have the central angle be $theta=frac {2pi}n$ so the only singularity is at $z=e^{pi i/n}$.

Integrating about the contour, we have$$begin{align*}ointlimits_{mathrm C}mathrm dz, f(z) & =intlimits_0^Rmathrm dx, f(x)+intlimits_{Gamma_R}mathrm dz, f(z)+intlimits_R^0mathrm dz, f(z)e^{pi i/n}\ & =(1-e^{pi i/n})intlimits_0^Rmathrm dx, f(x)+intlimits_{Gamma_R}mathrm dz, f(z)end{align*}$$

The arc integral vanishes as $R$ tends towards infinity. This can be verified using the estimation lemma$$left|,intlimits_{Gamma_R}mathrm dz, f(z),right|leq ML$$where $L$ is the length of the contour and $|f(z)|$ is bounded by a maximum $M$. The length of the arc can be easily calculated to be $L=tfrac {2pi R}n$. And through the triangle inequality, the max can be found$$|z|^n=left|z^nright|=left|z^n+1-1right|leqleft|z^n+1right|+1$$Since $|z|=R$, then we have that$$left|,intlimits_{Gamma_{R}}mathrm dz, f(z),right|leqfrac {1}{R^n-1}frac {2pi R}nxrightarrow{R,to,infty}0$$

Hence$$ointlimits_{mathrm C}mathrm dz, f(z)=(1-e^{pi i/n})intlimits_0^{infty}mathrm dx, f(x)$$

The contour integral is equal to $2pi i$ times the sum of its residues. Fortunately for us, due to our clever thinking in the beginning, we've managed to only encase one singularity of $f(z)$ within our contour: $z=e^{pi i/n}$. Therefore, the residue can be calculated as$$begin{align*}operatorname*{Res}_{z,=, e^{pi i/n}}f(z) & =limlimits_{zto e^{pi i/n}}frac {z-e^{pi i/n}}{1+z^n}\ & =limlimits_{zto e^{pi i/n}}frac z{nz^n}\ & =-frac {e^{pi i/n}}nend{align*}$$Note that above, I have used L Hopital's rule as a shortcut. Putting everything together$$begin{align*}intlimits_0^{infty}frac {mathrm dx}{1+x^n} & =-frac {2pi i}nfrac {e^{pi i/n}}{1-e^{2pi i/n}}\ & =frac {pi}nfrac {2i}{e^{pi i/n}-e^{-pi i/n}}\ & color{blue}{,=frac {pi}ncscleft(frac {pi}nright)}end{align*}$$

The integral under question is simply the case when $n=3$. Therefore$$mathfrak{I}(color{brown}{3})=intlimits_0^{infty}frac {mathrm dx}{1+x^{color{brown}{3}}}=frac {pi}{color{brown}{3}}cscleft(frac {pi}{color{brown}{3}}right)color{red}{=frac {2pi}{3sqrt3}}$$