# Evaluate the Integral $$\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}$$

How to evalute

\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}

was a question posed at Mathematics Stack Exchange. Here is my solution

The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and the lower limit of integration equal to the real root. So we have

\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(t-a)}{m+(t-a)},k \right)

where

m^{2} = (r – a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} – \frac{a-r}{2m}

$$\mathrm{cn}^{-1}z$$ is the inverse of one of the Jacobi elliptic functions and $$k$$ is the modulus.

To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}

We now have

a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} – \frac{3}{4\sqrt{3}}

and thus

\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 – x}{\sqrt{3} – 1 + x},\sqrt{\frac{1}{2} – \frac{3}{4\sqrt{3}}} \right)