How do i evaluate

$\displaystyle

\int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}}

$

It is part of a previous question where i prooved that the derivative of

$\displaystyle \frac{dy}{dx}sech^-1(x) = -\frac{1}{x\sqrt{x^2-1}}$

I'm sure that i have to do some rearranging of the problem to get it into the derivative of [tex] sech^-1(x) [\math].

I have tried to do the problem and so far the solution i have is

$\displaystyle \int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}} = \int_{1}^{x}dz\sqrt{\frac{z}{z^3}} + \int_{1}^{x}dz\sqrt{\frac{z}{-1}} = \int_{1}^{x}dz\sqrt{\frac{1}{z^2}} - \int_{1}^{x}dz\sqrt{z} $

Either i have made a mistake somewhere or i am going about the problem the wrong way but i have hit a brick wall. Can someone please give me some feedback on what i need to do or what i have done wrong