Evaluate $\int \frac{1}{x^3+3x+1}dx$ Evaluate $$\int \frac{1}{x^3+3x+1}dx$$
I tried to evaluate it but I couldn't  do . 

By setting $x=2z$ we have
$$I=\int\frac{dx}{x^3+3x+1}=\int\frac{dz}{4z^3+3z+1/2}$$
and by replacing $z$ with $\sinh t$, then $t$ with $\log u$,  we have:
$$ I = \int \frac{\cosh t}{1/2+\sinh(3t)}\,dt =\int\frac{u^3+u}{u^6+u^3-1}\,du$$
that is a bit better to deal with and, more important, gives that:
$$ 4z^3+3z+1/2 = \left(z+\sinh\frac{\operatorname{arcsinh} 1/2}{3}\right)\left(4z^2-4\sinh\frac{\operatorname{arcsinh} 1/2}{3}z+\frac{1}{2\sinh\frac{\operatorname{arcsinh} 1/2}{3}}\right)$$
so we have the roots of $x^3+3x+1$ in a explicit trigonometric form and we can compute $I$ through partial fraction decomposition. We also have:
$$\operatorname{arcsinh}\frac{1}{2}=\log\left(\frac{1+\sqrt{5}}{2}\right).$$
