<s>[INST] Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ $$\int^\infty_0\frac{1}{x^3+1}\,dx$$
The answer is $\frac{2\pi}{3\sqrt{3}}$.
How can I evaluate this integral?
 [/INST] Using  Partial Fraction Decomposition,
$$\frac1{1+x^3}=\frac A{1+x}+\frac{Bx+C}{1-x+x^2}$$
Multiply either sides by $(1+x)(1-x+x^2)$ and compare the coefficients of the different powers of $x$ to find $A,B,C$
As $\displaystyle1-x+x^2=\frac{4x^2-4x+4}4=\frac{(2x-1)^2+3}4$
Using Trigonometric substitution, set $\displaystyle 2x-1=\sqrt3\tan\phi$ in the second part
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